Each month Blake gets the report containing main economic indicators of the company "Blake Technologies". There are n commodities produced by the company. For each of them there is exactly one integer in the final report, that denotes corresponding revenue. Before the report gets to Blake, it passes through the hands of m managers. Each of them may reorder the elements in some order. Namely, the i-th manager either sorts first ri numbers in non-descending or non-ascending order and then passes the report to the manager i + 1, or directly to Blake (if this manager has number i = m).

Employees of the "Blake Technologies" are preparing the report right now. You know the initial sequence ai of length n and the description of each manager, that is value ri and his favourite order. You are asked to speed up the process and determine how the final report will look like.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 200 000) — the number of commodities in the report and the number of managers, respectively.

The second line contains n integers ai (|ai| ≤ 109) — the initial report before it gets to the first manager.

Then follow m lines with the descriptions of the operations managers are going to perform. The i-th of these lines contains two integers ti and ri (<image>, 1 ≤ ri ≤ n), meaning that the i-th manager sorts the first ri numbers either in the non-descending (if ti = 1) or non-ascending (if ti = 2) order.

Output

Print n integers — the final report, which will be passed to Blake by manager number m.

Examples

Input

3 1
1 2 3
2 2


Output

2 1 3 

Input

4 2
1 2 4 3
2 3
1 2


Output

2 4 1 3 

Note

In the first sample, the initial report looked like: 1 2 3. After the first manager the first two numbers were transposed: 2 1 3. The report got to Blake in this form.

In the second sample the original report was like this: 1 2 4 3. After the first manager the report changed to: 4 2 1 3. After the second manager the report changed to: 2 4 1 3. This report was handed over to Blake.
#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
long long fac[1000006];
inline long long exp(long long x, long long n) {
  long long r = 1;
  x = x;
  while (n) {
    if (n % 2) r = (r * x) % 1000000007;
    x = (x * x) % 1000000007;
    n = n / 2;
  }
  return r;
}
inline long long mmi(long long a) {
  return exp(a, 1000000007 - 2) % 1000000007;
}
inline long long fact(long long n) {
  long long res = 1;
  for (long long i = 1; i < (n + 1); ++i) {
    res = (res * i) % 1000000007;
  }
  return res;
}
inline void fact_a() {
  fac[0] = 1;
  fac[1] = 1;
  for (long long i = 1; i < (105); ++i) {
    fac[i] = (fac[i - 1] * i) % 1000000007;
  }
}
inline long long inv_fact(long long n) {
  long long par = fac[n];
  long long res = mmi(par);
  return res;
}
inline long long comb(long long n, long long r) {
  if (n == r && r == 0) return 1;
  if (n < r) return 0;
  return ((fac[n] * inv_fact(r)) % 1000000007 * inv_fact(n - r)) % 1000000007;
}
struct triplet {
  long long a, b, c;
};
bool operator<(const triplet &t1, const triplet &t2) {
  if (t1.a < t2.a) return true;
  if (t1.a == t2.a && t1.b < t2.b) return true;
  if (t1.a == t2.a && t1.b == t2.b && t1.c < t2.c) return true;
  return false;
}
pair<long long, pair<long long, long long> > ex_gcd(long long a, long long b) {
  if (b == 0) {
    return make_pair(a, make_pair(1, 0));
  }
  pair<long long, pair<long long, long long> > p = ex_gcd(b, a % b);
  long long gcd = p.first;
  long long x1 = p.second.first;
  long long y1 = p.second.second;
  long long x = y1;
  long long y = x1 - (a / b) * y1;
  return make_pair(gcd, make_pair(x, y));
}
long long prime[3000006];
long long spf_prime[3000006];
void sieve() {
  prime[1] = 1;
  for (long long i = 2; i * i <= 3000000; i++)
    if (prime[i] == 0)
      for (long long j = i * i; j <= 3000000; j += i) prime[j] = 1;
}
void spf() {
  for (long long i = 2; i * i <= 3000000; i++)
    if (!spf_prime[i])
      for (long long j = i * i; j <= 3000000; j += i)
        if (!spf_prime[j]) spf_prime[j] = i;
  for (long long i = 2; i <= 3000000; i++)
    if (!spf_prime[i]) spf_prime[i] = i;
}
long long getparent_BIT(long long idx) { return idx - (idx & -idx); }
long long getnext_BIT(long long idx) { return idx + (idx & -idx); }
long long getsum_BIT(long long idx, long long BIT[], long long n) {
  long long sum = 0;
  while (idx > 0) {
    sum += BIT[idx];
    idx = getparent_BIT(idx);
  }
  return sum;
}
void update_BIT(long long idx, long long BIT[], long long val, long long n) {
  while (idx <= n) {
    BIT[idx] += val;
    idx = getnext_BIT(idx);
  }
}
void build_BIT(long long BIT[], long long a[], long long n) {
  for (long long i = 0; i < (n); ++i) {
    update_BIT(i, BIT, a[i], n);
  }
}
void comp_lps(string s, long long lps[], long long n) {
  long long i = 1, j = 0;
  while (i < n) {
    if (s[i] == s[j]) {
      lps[i++] = ++j;
    } else {
      if (j != 0)
        j = lps[j - 1];
      else
        lps[i++] = 0;
    }
  }
}
void dfs_l(long long u, vector<vector<long long> > &adj, long long status[]) {
  if (status[u] == 1) return;
  status[u] = 1;
  for (long long i = 0; i < (adj[u].size()); ++i) dfs_l(adj[u][i], adj, status);
}
long long dij(long long u, long long n,
              vector<vector<pair<long long, long long> > > &adj,
              long long status[]) {
  long long dist[n];
  for (long long i = 0; i < (n); ++i) dist[i] = 1000000000;
  dist[0] = 0;
  priority_queue<pair<long long, long long>,
                 vector<pair<long long, long long> >,
                 greater<pair<long long, long long> > >
      q;
  q.push(make_pair(0, 0));
  while (!q.empty()) {
    pair<long long, long long> p;
    p = q.top();
    q.pop();
    long long x = p.second;
    long long w = p.first;
    if (status[x] == 1) continue;
    status[x] = 1;
    for (long long i = 0; i < (adj[x].size()); ++i) {
      long long ww = adj[x][i].first;
      long long y = adj[x][i].second;
      if (dist[x] + ww < dist[y]) {
        dist[y] = dist[x] + ww;
        q.push(make_pair(dist[y], y));
      }
    }
  }
  if (dist[n - 1] >= 1000000000) return -1;
  return dist[n - 1];
}
long long phi(long long n) {
  vector<long long> v;
  long long k = n;
  ;
  for (long long i = 2; i * i <= n; i++) {
    if (k % i == 0) {
      v.push_back(i);
      while (k % i == 0) k /= i;
    }
  }
  if (k > 1) v.push_back(k);
  long long ans = n;
  for (long long i = 0; i < (v.size()); ++i) ans -= (ans / v[i]);
  return ans;
}
struct trie {
  struct trie *left, *right;
  bool isend;
};
trie *get_trie() {
  trie *ptr = new trie;
  ptr->left = NULL;
  ptr->right = NULL;
  ptr->isend = false;
  return ptr;
}
bool isempty(trie *root) {
  if (root->left == NULL && root->right == NULL) return true;
  return false;
}
void insert_trie(trie *root, string key) {
  trie *ptr = root;
  for (long long i = 0; i < (key.size()); ++i) {
    if (key[i] == '0') {
      if (ptr->left == NULL) {
        ptr->left = get_trie();
      }
      ptr = ptr->left;
    } else {
      if (ptr->right == NULL) ptr->right = get_trie();
      ptr = ptr->right;
    }
  }
  ptr->isend = true;
}
trie *remove_trie(trie *root, string key, long long idx = 0) {
  if (root == NULL) return NULL;
  if (idx == key.size()) {
    if (isempty(root)) {
      delete (root);
      root = NULL;
    }
    return root;
  }
  if (key[idx] == '0')
    root->left = remove_trie(root->left, key, idx + 1);
  else
    root->right = remove_trie(root->right, key, idx + 1);
  if (isempty(root)) {
    delete (root);
    root = NULL;
  }
  return root;
}
long long search_trie(trie *root, string key) {
  trie *ptr = root;
  string s = "";
  for (long long i = 0; i < (key.size()); ++i) {
    if (key[i] == '0') {
      if (ptr->right != NULL) {
        ptr = ptr->right;
        s += '1';
      } else {
        ptr = ptr->left;
        s += '0';
      }
    } else {
      if (ptr->left != NULL) {
        ptr = ptr->left;
        s += '1';
      } else {
        ptr = ptr->right;
        s += '0';
      }
    }
  }
  long long res = 0;
  reverse(s.begin(), s.end());
  for (long long i = s.length() - 1; i >= (0); --i)
    res = res * 2 + (s[i] - '0');
  return res;
}
struct Comp {
  bool operator()(const std::pair<int, int> &a, const std::pair<int, int> &b) {
    if (a.first != b.first) {
      return a.first < b.first;
    }
    return a.second > b.second;
  }
};
void dearr() {
  long long dar[1005];
  dar[0] = 1;
  dar[1] = 0;
  dar[2] = 1;
  for (long long i = 3; i < (1002); ++i) {
    dar[i] = (i - 1) * (dar[i - 1] + dar[i - 2]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m;
  cin >> n >> m;
  long long a[n];
  for (long long i = 0; i < (n); ++i) cin >> a[i];
  stack<pair<long long, long long> > s;
  long long b[n];
  for (long long i = 0; i < (m); ++i) {
    long long t, r;
    cin >> t >> r;
    r--;
    while (!s.empty() && s.top().first <= r) {
      s.pop();
    }
    s.push(make_pair(r, t));
  }
  vector<pair<long long, long long> > v;
  while (!s.empty()) {
    v.push_back(s.top());
    s.pop();
  }
  sort(v.rbegin(), v.rend());
  vector<long long> u;
  long long val = n - 1 - v[0].first;
  for (long long i = 0; i < (n - val); ++i) u.push_back(a[i]);
  sort(u.begin(), u.end());
  for (long long i = n - val; i < (n); ++i) b[i] = a[i];
  for (long long i = 0; i < (u.size()); ++i) b[i] = u[i];
  long long ptr1 = 0;
  long long ptr2 = n - 1 - val;
  long long ptr3 = 0;
  long long ptr4 = n - 1 - val;
  long long last = -1;
  if (v[0].second == 1)
    last = 1;
  else
    last = 2;
  for (long long i = 1; i < (v.size()); ++i) {
    long long idx = v[i].first;
    long long op = v[i].second;
    long long num = v[i - 1].first - idx;
    if (last == 1) {
      while (num--) {
        a[ptr4] = b[ptr2];
        ptr2--;
        ptr4--;
      }
    } else {
      while (num--) {
        a[ptr4] = b[ptr1];
        ptr1++;
        ptr4--;
      }
    }
    last = op;
  }
  long long idx = v[v.size() - 1].first;
  vector<long long> y;
  for (long long i = ptr1; i < (ptr1 + idx + 1); ++i) y.push_back(b[i]);
  if (last == 1)
    sort(y.begin(), y.end());
  else
    sort(y.rbegin(), y.rend());
  for (long long i = 0; i < (idx + 1); ++i) a[i] = y[i];
  for (auto i : a) cout << i << " ";
  cout << "\n";
  ;
}
