You are given a connected weighted graph with n vertices and m edges. The graph doesn't contain loops nor multiple edges. Consider some edge with id i. Let's determine for this edge the maximum integer weight we can give to it so that it is contained in all minimum spanning trees of the graph if we don't change the other weights.

You are to determine this maximum weight described above for each edge. You should calculate the answer for each edge independently, it means there can't be two edges with changed weights at the same time.

Input

The first line contains two integers n and m (2 ≤ n ≤ 2·105, n - 1 ≤ m ≤ 2·105), where n and m are the number of vertices and the number of edges in the graph, respectively.

Each of the next m lines contains three integers u, v and c (1 ≤ v, u ≤ n, v ≠ u, 1 ≤ c ≤ 109) meaning that there is an edge between vertices u and v with weight c. 

Output

Print the answer for each edge in the order the edges are given in the input. If an edge is contained in every minimum spanning tree with any weight, print -1 as the answer.

Examples

Input

4 4
1 2 2
2 3 2
3 4 2
4 1 3


Output

2 2 2 1 

Input

4 3
1 2 2
2 3 2
3 4 2


Output

-1 -1 -1 
#include <bits/stdc++.h>
using namespace std;
const int K = 2;
const double PI = acos(-1);
template <class T>
inline T bigmod(T a, T b, T M) {
  long long ret = 1;
  for (; b > 0; b >>= 1) {
    if (b & 1) ret = (ret * a) % M;
    a = (a * a) % M;
  }
  return ret;
}
template <class T>
inline T gcd(T a, T b) {
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
template <class T>
inline T lcm(T a, T b) {
  return (a / gcd(a, b)) * b;
}
template <class T>
inline T poww(T a, T b) {
  long long ret = 1;
  for (; b > 0; b >>= 1) {
    if (b & 1) ret = ret * a;
    a = a * a;
  }
  return (T)ret;
}
pair<int, int> exEuclid(int a, int b) {
  if (b == 0)
    return pair<int, int>(1, 0);
  else {
    pair<int, int> r = exEuclid(b, a % b);
    return pair<int, int>(r.second, r.first - r.second * (a / b));
  }
}
int modInv(int a, int m) {
  pair<int, int> ret = exEuclid(a, m);
  return ((ret.first % m) + m) % m;
}
int Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
int n, m, x, y, z;
struct edge {
  int u, v, id, w;
  edge(int _u, int _v, int _w, int _id) {
    u = _u;
    v = _v;
    w = _w;
    id = _id;
  }
};
bool comp(edge a, edge b) { return a.w < b.w; }
vector<edge> vec, nvec;
int par[2 * 100010], pa[2 * 100010][40], del[2 * 100010], rnk[2 * 100010],
    mark[2 * 100010], tmp[2 * 100010], ans[2 * 100010], depth[2 * 100010],
    mx[2 * 100010][40], sub[2 * 100010];
vector<pair<int, pair<int, int> > > tree[2 * 100010], nontree[2 * 100010];
int find(int u) {
  if (par[u] == u) return u;
  return par[u] = find(par[u]);
}
void dfs(int u, int from, int dep, int we) {
  depth[u] = dep;
  pa[u][0] = from;
  mx[u][0] = we;
  for (int i = (0); i < (tree[u].size()); i++) {
    int v = tree[u][i].first;
    if (from == v) continue;
    dfs(v, u, dep + 1, tree[u][i].second.first);
  }
}
int LCA(int p, int q) {
  if (depth[q] > depth[p]) swap(p, q);
  int lg = 0;
  while (1) {
    int next = lg + 1;
    if ((1 << next) > depth[p]) break;
    lg++;
  }
  for (int i = lg; i >= 0; i--) {
    if (depth[p] - (1 << i) >= depth[q]) p = pa[p][i];
  }
  if (p == q) return p;
  for (int i = lg; i >= 0; i--) {
    if (pa[p][i] != -1 && pa[p][i] != pa[q][i]) {
      p = pa[p][i];
      q = pa[q][i];
    }
  }
  return pa[p][0];
}
int getmax(int p, int q) {
  int ret = -1;
  int lg = 0;
  if (p == q) return ret;
  int dif = depth[p] - depth[q];
  while (1) {
    int next = lg + 1;
    if ((1 << next) > dif) break;
    lg++;
  }
  for (int i = lg; i >= 0; i--) {
    if (depth[p] - (1 << i) >= depth[q]) {
      ret = max(ret, mx[p][i]);
      p = pa[p][i];
    }
  }
  return ret;
}
void outoftree() {
  memset(pa, -1, sizeof pa);
  dfs(1, -1, 0, -1);
  for (int j = 1; 1 << j <= n; j++) {
    for (int i = 1; i <= n; i++) {
      if (pa[i][j - 1] != -1) {
        pa[i][j] = pa[pa[i][j - 1]][j - 1];
        mx[i][j] = max(mx[pa[i][j - 1]][j - 1], mx[i][j - 1]);
      }
    }
  }
  for (int i = (0); i < (nvec.size()); i++) {
    int lca = LCA(nvec[i].u, nvec[i].v);
    ans[nvec[i].id] = max(getmax(nvec[i].u, lca), getmax(nvec[i].v, lca)) - 1;
  }
}
int nn;
vector<int> va;
void dfs1(int u, int from) {
  sub[u] = 1;
  nn++;
  for (int i = (0); i < (tree[u].size()); i++) {
    int v = tree[u][i].first;
    if (v == from || del[v]) continue;
    dfs1(v, u);
    sub[u] += sub[v];
  }
}
int dfs2(int u, int from) {
  for (int i = (0); i < (tree[u].size()); i++) {
    int v = tree[u][i].first;
    if (v == from || del[v]) continue;
    if (sub[v] > nn / 2) return dfs2(v, u);
  }
  return u;
}
void dfs3(int u, int from, int tt) {
  mark[u] = tt;
  tmp[u] = 1061109567;
  va.push_back(u);
  for (int i = (0); i < (tree[u].size()); i++) {
    int v = tree[u][i].first;
    if (v == from || del[v]) continue;
    dfs3(v, u, tt);
  }
}
void dfs4(int u, int from, int idx) {
  for (int i = (0); i < (tree[u].size()); i++) {
    int v = tree[u][i].first;
    if (v == from || del[v]) continue;
    dfs4(v, u, tree[u][i].second.second);
    tmp[u] = min(tmp[u], tmp[v]);
  }
  if (from != -1 && tmp[u] != 1061109567) {
    ans[idx] = min(ans[idx], tmp[u] - 1);
  }
}
void decompose(int root, int from) {
  nn = 0;
  dfs1(root, root);
  int centriod = dfs2(root, root);
  del[centriod] = 1;
  va.clear();
  int tt = 1;
  mark[centriod] = tt;
  tmp[centriod] = 1061109567;
  va.push_back(centriod);
  for (int i = (0); i < (tree[centriod].size()); i++) {
    int v = tree[centriod][i].first;
    if (del[v]) continue;
    tt++;
    dfs3(v, centriod, tt);
  }
  for (int i = (0); i < (va.size()); i++) {
    int u = va[i];
    for (int j = (0); j < (nontree[u].size()); j++) {
      int v = nontree[u][j].first;
      if (mark[u] && mark[v] && mark[u] != mark[v]) {
        tmp[u] = min(tmp[u], nontree[u][j].second.first);
        tmp[v] = min(tmp[v], nontree[u][j].second.first);
      }
    }
  }
  dfs4(centriod, -1, -1);
  for (int i = (0); i < (va.size()); i++) mark[va[i]] = 0;
  for (int i = (0); i < (tree[centriod].size()); i++) {
    int v = tree[centriod][i].first;
    if (del[v]) continue;
    decompose(v, centriod);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = (1); i <= (m); i++) {
    scanf("%d %d %d", &x, &y, &z);
    vec.push_back(edge(x, y, z, i));
  }
  sort((vec).begin(), (vec).end(), comp);
  for (int i = (1); i <= (n); i++) par[i] = i;
  for (int i = (0); i < (m); i++) {
    int uu = vec[i].u, vv = vec[i].v, ww = vec[i].w, idx = vec[i].id;
    int pu = find(uu);
    int pv = find(vv);
    if (pu != pv) {
      par[pu] = pv;
      tree[uu].push_back(make_pair(vv, pair<int, int>(ww, idx)));
      tree[vv].push_back(make_pair(uu, pair<int, int>(ww, idx)));
    } else {
      nontree[uu].push_back(make_pair(vv, pair<int, int>(ww, idx)));
      nontree[vv].push_back(make_pair(uu, pair<int, int>(ww, idx)));
      nvec.push_back(edge(uu, vv, ww, idx));
    }
  }
  memset(ans, 63, sizeof ans);
  outoftree();
  decompose(1, -1);
  for (int i = (1); i <= (m); i++) {
    if (ans[i] == 1061109567) ans[i] = -1;
    printf("%d ", ans[i]);
  }
  printf("\n");
  return 0;
}
