In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation 

F1 = 1; F2 = 1; Fn = Fn - 1 + Fn - 2 (n > 2).

DZY loves Fibonacci numbers very much. Today DZY gives you an array consisting of n integers: a1, a2, ..., an. Moreover, there are m queries, each query has one of the two types:

  1. Format of the query "1 l r". In reply to the query, you need to add Fi - l + 1 to each element ai, where l ≤ i ≤ r. 
  2. Format of the query "2 l r". In reply to the query you should output the value of <image> modulo 1000000009 (109 + 9). 



Help DZY reply to all the queries.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 300000). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — initial array a.

Then, m lines follow. A single line describes a single query in the format given in the statement. It is guaranteed that for each query inequality 1 ≤ l ≤ r ≤ n holds.

Output

For each query of the second type, print the value of the sum on a single line.

Examples

Input

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


Output

17
12

Note

After the first query, a = [2, 3, 5, 7].

For the second query, sum = 2 + 3 + 5 + 7 = 17.

After the third query, a = [2, 4, 6, 9].

For the fourth query, sum = 2 + 4 + 6 = 12.
#include <bits/stdc++.h>
using namespace std;
long long f[300005], s[300005], A[300005], tree[1200000];
pair<long long, long long> lazy[1200000];
long long getFib(long long a, long long b, int n) {
  if (n == 1) return a;
  if (n == 2) return b;
  return (a * f[n - 2] + b * f[n - 1]) % 1000000009;
}
long long getSum(long long a, long long b, int n) {
  if (n == 1) return a;
  if (n == 2) return (a + b) % 1000000009;
  return (a * (s[n - 2] + 1) + b * s[n - 1]) % 1000000009;
}
void build(int pos, int l, int r) {
  if (l == r) {
    tree[pos] = A[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(pos * 2, l, mid);
  build(pos * 2 + 1, mid + 1, r);
  tree[pos] = (tree[pos * 2] + tree[pos * 2 + 1]) % 1000000009;
}
void update(int pos, int l, int r, int a, int b) {
  int mid = (l + r) >> 1;
  if (lazy[pos] != make_pair(0LL, 0LL)) {
    tree[pos] =
        (tree[pos] + getSum(lazy[pos].first, lazy[pos].second, r - l + 1)) %
        1000000009;
    if (l != r) {
      lazy[pos * 2].first =
          (lazy[pos * 2].first + lazy[pos].first) % 1000000009;
      lazy[pos * 2].second =
          (lazy[pos * 2].second + lazy[pos].second) % 1000000009;
      lazy[pos * 2 + 1].first =
          (lazy[pos * 2 + 1].first +
           getFib(lazy[pos].first, lazy[pos].second, mid + 1 - l + 1)) %
          1000000009;
      lazy[pos * 2 + 1].second =
          (lazy[pos * 2 + 1].second +
           getFib(lazy[pos].first, lazy[pos].second, mid + 2 - l + 1)) %
          1000000009;
    }
    lazy[pos] = {0, 0};
  }
  if (l > b or r < a) return;
  if (l >= a and r <= b) {
    tree[pos] = (tree[pos] + s[r - a + 1] - s[l - a] + 1000000009) % 1000000009;
    if (l != r) {
      lazy[pos * 2].first = (lazy[pos * 2].first + f[l - a + 1]) % 1000000009;
      lazy[pos * 2].second = (lazy[pos * 2].second + f[l - a + 2]) % 1000000009;
      lazy[pos * 2 + 1].first =
          (lazy[pos * 2 + 1].first + f[mid + 1 - a + 1]) % 1000000009;
      lazy[pos * 2 + 1].second =
          (lazy[pos * 2 + 1].second + f[mid + 2 - a + 1]) % 1000000009;
    }
    return;
  }
  update(pos * 2, l, mid, a, b);
  update(pos * 2 + 1, mid + 1, r, a, b);
  tree[pos] = (tree[pos * 2] + tree[pos * 2 + 1]) % 1000000009;
}
long long query(int pos, int l, int r, int a, int b) {
  int mid = (l + r) >> 1;
  if (lazy[pos] != make_pair(0LL, 0LL)) {
    tree[pos] =
        (tree[pos] + getSum(lazy[pos].first, lazy[pos].second, r - l + 1)) %
        1000000009;
    if (l != r) {
      lazy[pos * 2].first =
          (lazy[pos * 2].first + lazy[pos].first) % 1000000009;
      lazy[pos * 2].second =
          (lazy[pos * 2].second + lazy[pos].second) % 1000000009;
      lazy[pos * 2 + 1].first =
          (lazy[pos * 2 + 1].first +
           getFib(lazy[pos].first, lazy[pos].second, mid + 1 - l + 1)) %
          1000000009;
      lazy[pos * 2 + 1].second =
          (lazy[pos * 2 + 1].second +
           getFib(lazy[pos].first, lazy[pos].second, mid + 2 - l + 1)) %
          1000000009;
    }
    lazy[pos] = {0, 0};
  }
  if (l > b or r < a) return 0;
  if (l >= a and r <= b) return tree[pos];
  return (query(pos * 2, l, mid, a, b) + query(pos * 2 + 1, mid + 1, r, a, b)) %
         1000000009;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, i, t, l, r;
  f[1] = f[2] = 1;
  for (i = 3; i <= 300000; i++) f[i] = (f[i - 1] + f[i - 2]) % 1000000009;
  for (i = 1; i <= 300000; i++) s[i] = (s[i - 1] + f[i]) % 1000000009;
  cin >> n >> m;
  for (i = 1; i <= n; i++) cin >> A[i];
  build(1, 1, n);
  while (m--) {
    cin >> t >> l >> r;
    if (t == 1)
      update(1, 1, n, l, r);
    else
      cout << query(1, 1, n, l, r) << '\n';
  }
  return 0;
}
