#include <bits/stdc++.h>
using namespace std;
const int N = (int)2e5 + 10;
const int P = (int)1e9;
int n, q, a[N];
long long fib[N * 2], *f;
int ft[2][N];
void Update(int *ft, int i, int v) {
  while (i <= n) {
    ft[i] = (ft[i] + v) % P;
    i += i & -i;
  }
}
int Query(int *ft, int l, int r) {
  int ans = 0;
  l--;
  for (; l; l -= l & -l) ans = (ans + P - ft[l]) % P;
  for (; r; r -= r & -r) ans = (ans + ft[r]) % P;
  return ans;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  f = fib + n;
  f[0] = 0;
  f[1] = 1;
  for (int i = -1; i >= -n; --i) f[i] = (f[i + 2] - f[i + 1] + P) % P;
  for (int i = 2; i <= n + 1; ++i) f[i] = (f[i - 1] + f[i - 2]) % P;
  for (int i = 1; i <= n; ++i) Update(ft[0], i, a[i] * f[i + 1] % P);
  for (int i = 1; i <= n; ++i) Update(ft[1], i, a[i] * f[i] % P);
  while (q--) {
    int k;
    scanf("%d", &k);
    if (k == 1) {
      int i, v;
      scanf("%d%d", &i, &v);
      Update(ft[0], i, (v + P - a[i]) * f[i + 1] % P);
      Update(ft[1], i, (v + P - a[i]) * f[i] % P);
      a[i] = v;
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      int x = Query(ft[0], l, r);
      int y = Query(ft[1], l, r);
      printf("%d\n", (int)((x * f[1 - l] + y * f[-l]) % P));
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

By the age of three Smart Beaver mastered all arithmetic operations and got this summer homework from the amazed teacher:

You are given a sequence of integers a1, a2, ..., an. Your task is to perform on it m consecutive operations of the following type:

  1. For given numbers xi and vi assign value vi to element axi. 
  2. For given numbers li and ri you've got to calculate sum <image>, where f0 = f1 = 1 and at i ≥ 2: fi = fi - 1 + fi - 2. 
  3. For a group of three numbers li ri di you should increase value ax by di for all x (li ≤ x ≤ ri). 



Smart Beaver planned a tour around great Canadian lakes, so he asked you to help him solve the given problem.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 2·105) — the number of integers in the sequence and the number of operations, correspondingly. The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 105). Then follow m lines, each describes an operation. Each line starts with an integer ti (1 ≤ ti ≤ 3) — the operation type: 

  * if ti = 1, then next follow two integers xi vi (1 ≤ xi ≤ n, 0 ≤ vi ≤ 105); 
  * if ti = 2, then next follow two integers li ri (1 ≤ li ≤ ri ≤ n); 
  * if ti = 3, then next follow three integers li ri di (1 ≤ li ≤ ri ≤ n, 0 ≤ di ≤ 105). 



The input limits for scoring 30 points are (subproblem E1): 

  * It is guaranteed that n does not exceed 100, m does not exceed 10000 and there will be no queries of the 3-rd type. 



The input limits for scoring 70 points are (subproblems E1+E2): 

  * It is guaranteed that there will be queries of the 1-st and 2-nd type only. 



The input limits for scoring 100 points are (subproblems E1+E2+E3): 

  * No extra limitations. 

Output

For each query print the calculated sum modulo 1000000000 (109).

Examples

Input

5 5
1 3 1 2 4
2 1 4
2 1 5
2 2 4
1 3 10
2 1 5


Output

12
32
8
50


Input

5 4
1 3 1 2 4
3 1 4 1
2 2 4
1 2 10
2 1 5


Output

12
45

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = (int)2e5 + 10;
const int P = (int)1e9;
int n, q, a[N];
long long fib[N * 2], *f;
int ft[2][N];
void Update(int *ft, int i, int v) {
  while (i <= n) {
    ft[i] = (ft[i] + v) % P;
    i += i & -i;
  }
}
int Query(int *ft, int l, int r) {
  int ans = 0;
  l--;
  for (; l; l -= l & -l) ans = (ans + P - ft[l]) % P;
  for (; r; r -= r & -r) ans = (ans + ft[r]) % P;
  return ans;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  f = fib + n;
  f[0] = 0;
  f[1] = 1;
  for (int i = -1; i >= -n; --i) f[i] = (f[i + 2] - f[i + 1] + P) % P;
  for (int i = 2; i <= n + 1; ++i) f[i] = (f[i - 1] + f[i - 2]) % P;
  for (int i = 1; i <= n; ++i) Update(ft[0], i, a[i] * f[i + 1] % P);
  for (int i = 1; i <= n; ++i) Update(ft[1], i, a[i] * f[i] % P);
  while (q--) {
    int k;
    scanf("%d", &k);
    if (k == 1) {
      int i, v;
      scanf("%d%d", &i, &v);
      Update(ft[0], i, (v + P - a[i]) * f[i + 1] % P);
      Update(ft[1], i, (v + P - a[i]) * f[i] % P);
      a[i] = v;
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      int x = Query(ft[0], l, r);
      int y = Query(ft[1], l, r);
      printf("%d\n", (int)((x * f[1 - l] + y * f[-l]) % P));
    }
  }
  return 0;
}

```