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;
template <class T>
inline T min(T &a, T &b) {
  return a < b ? a : b;
}
template <class T>
inline T max(T &a, T &b) {
  return a > b ? a : b;
}
template <class T>
void read(T &x) {
  char ch;
  while ((ch = getchar()) && !isdigit(ch))
    ;
  x = ch - '0';
  while ((ch = getchar()) && isdigit(ch)) x = x * 10 + ch - '0';
}
struct point {
  int x, y;
  point() {}
  point(int _x, int _y) : x(_x), y(_y) {}
};
const int N = 400000;
const int Mod = 1000000009;
int n, fib[N];
struct lj {
  int t[N];
  void add(int x, int c) {
    for (int i = x; i <= n; i += i & (-i)) t[i] = (t[i] + c) % Mod;
  }
  int ask(int x) {
    int ans = 0;
    for (int i = x; i; i -= i & (-i)) ans = (ans + t[i]) % Mod;
    return ans;
  }
} a, b, c, d;
int Chen(int a, int b) {
  return int((long long)a * (long long)b % (long long)Mod);
}
int ask(int x) {
  if (x == 0) return 0;
  long long ans = (long long)Chen(a.ask(x), fib[x + 2]) - (long long)b.ask(x) -
                  (long long)Chen(c.ask(x), fib[x + 3]) + (long long)d.ask(x);
  ans = ans % (long long)Mod;
  if (ans < 0) ans += Mod;
  return (int)ans;
}
int m, l, r, x, ty, sum[N];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &x), sum[i] = (sum[i - 1] + x) % Mod;
  fib[1] = 1;
  fib[2] = 1;
  for (int i = 3; i <= n + 10; i++) fib[i] = (fib[i - 1] + fib[i - 2]) % Mod;
  n++;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &ty, &l, &r);
    if (ty == 1) {
      a.add(l, l & 1 ? fib[l] : -fib[l]);
      a.add(r + 1, l & 1 ? -fib[l] : fib[l]);
      b.add(l, l & 1 ? Chen(fib[l], fib[l + 1]) : -Chen(fib[l], fib[l + 1]));
      b.add(r + 1,
            l & 1 ? -Chen(fib[l], fib[r + 2]) : Chen(fib[l], fib[r + 2]));
      c.add(l, l & 1 ? fib[l - 1] : -fib[l - 1]);
      c.add(r + 1, l & 1 ? -fib[l - 1] : fib[l - 1]);
      d.add(l, l & 1 ? Chen(fib[l - 1], fib[l + 2])
                     : -Chen(fib[l - 1], fib[l + 2]));
      d.add(r + 1, l & 1 ? -Chen(fib[l - 1], fib[r + 3])
                         : Chen(fib[l - 1], fib[r + 3]));
    } else {
      int t1 = (sum[r] - sum[l - 1] + Mod) % Mod,
          t2 = (ask(r) - ask(l - 1) + Mod) % Mod;
      printf("%d\n", (t1 + t2) % Mod);
    }
  }
}
