#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000000;
struct fib {
  long long a, b, c;
  inline fib() { a = b = c = 0; }
  inline fib(long long x, long long y, long long z) {
    a = x % MOD, b = y % MOD, c = z % MOD;
  }
  inline fib operator+(const fib &x) const {
    return fib(a + x.a, b + x.b, c + x.c);
  }
  inline fib operator*(const fib &x) const {
    return fib(a * x.a + b * x.b, a * x.b + b * x.c, b * x.b + c * x.c);
  }
  inline fib operator*(const int x) const { return fib(a * x, b * x, c * x); }
  inline fib operator+=(const fib &x) { return *this = *this + x; }
  inline fib operator*=(const fib &x) { return *this = *this * x; }
  inline fib operator*=(const int x) { return *this = *this * x; }
} base(0, 1, 1), pw[233333];
struct node {
  int l, r, lazy;
  fib val, par;
  node *lson, *rson;
  inline void upd2() { par = rson->par * pw[lson->r - lson->l] + lson->par; }
  inline void upd() { val = rson->val * pw[lson->r - lson->l] + lson->val; }
  inline void mark(int x) {
    lazy += x;
    val += par * x;
  }
  inline void fa_lazy() {
    if (lazy) {
      lson->mark(lazy);
      rson->mark(lazy);
      lazy = 0;
    }
  }
} * root, pool[555555];
int n, m, top;
int v[233333];
node *build(int l, int r) {
  node *tmp = &pool[top++];
  tmp->l = l;
  tmp->r = r;
  if (l != r) {
    int mid = (l + r) >> 1;
    tmp->lson = build(l, mid);
    tmp->rson = build(mid + 1, r);
    tmp->upd2();
    tmp->upd();
  } else {
    tmp->par = base;
    tmp->val = tmp->par * v[l];
  }
  return tmp;
}
void change(node *cur, int pos, int x) {
  if (cur->l == cur->r) {
    cur->val = cur->par * x;
    return;
  }
  if (pos <= cur->lson->r)
    change(cur->lson, pos, x);
  else
    change(cur->rson, pos, x);
  cur->upd();
}
void add(node *cur, int l, int r, int x) {
  if (cur->l == l && cur->r == r) {
    cur->mark(x);
    return;
  }
  cur->fa_lazy();
  if (cur->lson->r >= r)
    add(cur->lson, l, r, x);
  else if (cur->rson->l <= l)
    add(cur->rson, l, r, x);
  else
    add(cur->lson, l, cur->lson->r, x), add(cur->rson, cur->rson->l, r, x);
  cur->upd();
}
fib query(node *cur, int l, int r) {
  if (cur->l == l && cur->r == r) return cur->val;
  cur->fa_lazy();
  if (cur->lson->r >= r)
    return query(cur->lson, l, r);
  else if (cur->rson->l <= l)
    return query(cur->rson, l, r);
  else
    return query(cur->lson, l, cur->lson->r) +
           query(cur->rson, cur->rson->l, r) * pw[cur->lson->r - l];
}
int main() {
  scanf("%d%d", &n, &m);
  pw[0] = base;
  for (int i = 1; i <= n; i++) pw[i] = pw[i - 1] * base;
  for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
  root = build(1, n);
  int ty, a, b, c;
  while (m--) {
    scanf("%d%d%d", &ty, &a, &b);
    if (ty == 1) {
      change(root, a, b);
    } else if (ty == 2) {
      fib tmp = query(root, a, b);
      printf("%lld\n", tmp.b);
    } else {
      scanf("%d", &c);
      add(root, a, b, c);
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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 MOD = 1000000000;
struct fib {
  long long a, b, c;
  inline fib() { a = b = c = 0; }
  inline fib(long long x, long long y, long long z) {
    a = x % MOD, b = y % MOD, c = z % MOD;
  }
  inline fib operator+(const fib &x) const {
    return fib(a + x.a, b + x.b, c + x.c);
  }
  inline fib operator*(const fib &x) const {
    return fib(a * x.a + b * x.b, a * x.b + b * x.c, b * x.b + c * x.c);
  }
  inline fib operator*(const int x) const { return fib(a * x, b * x, c * x); }
  inline fib operator+=(const fib &x) { return *this = *this + x; }
  inline fib operator*=(const fib &x) { return *this = *this * x; }
  inline fib operator*=(const int x) { return *this = *this * x; }
} base(0, 1, 1), pw[233333];
struct node {
  int l, r, lazy;
  fib val, par;
  node *lson, *rson;
  inline void upd2() { par = rson->par * pw[lson->r - lson->l] + lson->par; }
  inline void upd() { val = rson->val * pw[lson->r - lson->l] + lson->val; }
  inline void mark(int x) {
    lazy += x;
    val += par * x;
  }
  inline void fa_lazy() {
    if (lazy) {
      lson->mark(lazy);
      rson->mark(lazy);
      lazy = 0;
    }
  }
} * root, pool[555555];
int n, m, top;
int v[233333];
node *build(int l, int r) {
  node *tmp = &pool[top++];
  tmp->l = l;
  tmp->r = r;
  if (l != r) {
    int mid = (l + r) >> 1;
    tmp->lson = build(l, mid);
    tmp->rson = build(mid + 1, r);
    tmp->upd2();
    tmp->upd();
  } else {
    tmp->par = base;
    tmp->val = tmp->par * v[l];
  }
  return tmp;
}
void change(node *cur, int pos, int x) {
  if (cur->l == cur->r) {
    cur->val = cur->par * x;
    return;
  }
  if (pos <= cur->lson->r)
    change(cur->lson, pos, x);
  else
    change(cur->rson, pos, x);
  cur->upd();
}
void add(node *cur, int l, int r, int x) {
  if (cur->l == l && cur->r == r) {
    cur->mark(x);
    return;
  }
  cur->fa_lazy();
  if (cur->lson->r >= r)
    add(cur->lson, l, r, x);
  else if (cur->rson->l <= l)
    add(cur->rson, l, r, x);
  else
    add(cur->lson, l, cur->lson->r, x), add(cur->rson, cur->rson->l, r, x);
  cur->upd();
}
fib query(node *cur, int l, int r) {
  if (cur->l == l && cur->r == r) return cur->val;
  cur->fa_lazy();
  if (cur->lson->r >= r)
    return query(cur->lson, l, r);
  else if (cur->rson->l <= l)
    return query(cur->rson, l, r);
  else
    return query(cur->lson, l, cur->lson->r) +
           query(cur->rson, cur->rson->l, r) * pw[cur->lson->r - l];
}
int main() {
  scanf("%d%d", &n, &m);
  pw[0] = base;
  for (int i = 1; i <= n; i++) pw[i] = pw[i - 1] * base;
  for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
  root = build(1, n);
  int ty, a, b, c;
  while (m--) {
    scanf("%d%d%d", &ty, &a, &b);
    if (ty == 1) {
      change(root, a, b);
    } else if (ty == 2) {
      fib tmp = query(root, a, b);
      printf("%lld\n", tmp.b);
    } else {
      scanf("%d", &c);
      add(root, a, b, c);
    }
  }
  return 0;
}

```