#include <bits/stdc++.h>
using namespace std;
long long arr[200005];
long long fib[200005], MOD = 1000000000;
int n, m;
long long tree1[(200005 << 2)], tree2[(200005 << 2)];
int lChild(int ind) { return (ind << 1) + 1; }
int rChild(int ind) { return (1 + ind) << 1; }
void init(long long *tree, long long *coeff, int ind, int left, int right) {
  int mid = (left + right) >> 1, lc = lChild(ind), rc = rChild(ind);
  if (left == right) {
    tree[ind] = (arr[left] * coeff[left]) % MOD;
    return;
  }
  init(tree, coeff, lc, left, mid);
  init(tree, coeff, rc, mid + 1, right);
  tree[ind] = tree[lc] + tree[rc];
  if (tree[ind] >= MOD) tree[ind] -= MOD;
}
int lo, hi;
void update1(long long *tree, long long *coeff, int ind, int left, int right,
             int val) {
  int mid = (left + right) >> 1, lc = lChild(ind), rc = rChild(ind);
  if (left == right) {
    tree[ind] = (val * coeff[left]) % MOD;
    return;
  }
  if (lo <= mid) update1(tree, coeff, lc, left, mid, val);
  if (hi > mid) update1(tree, coeff, rc, mid + 1, right, val);
  tree[ind] = tree[lc] + tree[rc];
  if (tree[ind] >= MOD) tree[ind] -= MOD;
}
long long query(long long *tree, long long *coeff, int ind, int left,
                int right) {
  int mid = (left + right) >> 1, lc = lChild(ind), rc = rChild(ind);
  if (lo > right || hi < left) return 0;
  if (lo <= left && right <= hi) return tree[ind];
  return (query(tree, coeff, lc, left, mid) +
          query(tree, coeff, rc, mid + 1, right)) %
         MOD;
}
int main() {
  fib[0] = fib[1] = 1;
  for (int i = 2; i < 200005; i++) {
    fib[i] = (fib[i - 1] + fib[i - 2]);
    if (fib[i] >= MOD) fib[i] -= MOD;
  }
  while (cin >> n >> m) {
    for (int i = 1; i <= n; i++) scanf("%I64d", &arr[i]);
    init(tree1, fib, 0, 1, n);
    init(tree2, fib + 1, 0, 1, n);
    while (m--) {
      int type, xx, v;
      scanf("%d", &type);
      if (type == 1) {
        scanf("%d%d", &xx, &v);
        arr[xx] = v;
        lo = hi = xx;
        update1(tree1, fib, 0, 1, n, v);
        update1(tree2, fib + 1, 0, 1, n, v);
      } else if (type == 2) {
        scanf("%d%d", &lo, &hi);
        long long sm = 0;
        sm = (query(tree2, fib + 1, 0, 1, n) * fib[lo - 1] -
              query(tree1, fib, 0, 1, n) * fib[lo]) %
             MOD;
        if (!(lo & 1)) sm = -sm;
        printf("%I64d\n", (sm + MOD) % MOD);
      }
    }
  }
  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;
long long arr[200005];
long long fib[200005], MOD = 1000000000;
int n, m;
long long tree1[(200005 << 2)], tree2[(200005 << 2)];
int lChild(int ind) { return (ind << 1) + 1; }
int rChild(int ind) { return (1 + ind) << 1; }
void init(long long *tree, long long *coeff, int ind, int left, int right) {
  int mid = (left + right) >> 1, lc = lChild(ind), rc = rChild(ind);
  if (left == right) {
    tree[ind] = (arr[left] * coeff[left]) % MOD;
    return;
  }
  init(tree, coeff, lc, left, mid);
  init(tree, coeff, rc, mid + 1, right);
  tree[ind] = tree[lc] + tree[rc];
  if (tree[ind] >= MOD) tree[ind] -= MOD;
}
int lo, hi;
void update1(long long *tree, long long *coeff, int ind, int left, int right,
             int val) {
  int mid = (left + right) >> 1, lc = lChild(ind), rc = rChild(ind);
  if (left == right) {
    tree[ind] = (val * coeff[left]) % MOD;
    return;
  }
  if (lo <= mid) update1(tree, coeff, lc, left, mid, val);
  if (hi > mid) update1(tree, coeff, rc, mid + 1, right, val);
  tree[ind] = tree[lc] + tree[rc];
  if (tree[ind] >= MOD) tree[ind] -= MOD;
}
long long query(long long *tree, long long *coeff, int ind, int left,
                int right) {
  int mid = (left + right) >> 1, lc = lChild(ind), rc = rChild(ind);
  if (lo > right || hi < left) return 0;
  if (lo <= left && right <= hi) return tree[ind];
  return (query(tree, coeff, lc, left, mid) +
          query(tree, coeff, rc, mid + 1, right)) %
         MOD;
}
int main() {
  fib[0] = fib[1] = 1;
  for (int i = 2; i < 200005; i++) {
    fib[i] = (fib[i - 1] + fib[i - 2]);
    if (fib[i] >= MOD) fib[i] -= MOD;
  }
  while (cin >> n >> m) {
    for (int i = 1; i <= n; i++) scanf("%I64d", &arr[i]);
    init(tree1, fib, 0, 1, n);
    init(tree2, fib + 1, 0, 1, n);
    while (m--) {
      int type, xx, v;
      scanf("%d", &type);
      if (type == 1) {
        scanf("%d%d", &xx, &v);
        arr[xx] = v;
        lo = hi = xx;
        update1(tree1, fib, 0, 1, n, v);
        update1(tree2, fib + 1, 0, 1, n, v);
      } else if (type == 2) {
        scanf("%d%d", &lo, &hi);
        long long sm = 0;
        sm = (query(tree2, fib + 1, 0, 1, n) * fib[lo - 1] -
              query(tree1, fib, 0, 1, n) * fib[lo]) %
             MOD;
        if (!(lo & 1)) sm = -sm;
        printf("%I64d\n", (sm + MOD) % MOD);
      }
    }
  }
  return 0;
}

```