Problem: At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:

  1. Print operation l, r. Picks should write down the value of <image>. 
  2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r). 
  3. Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x). 



Can you help Picks to perform the whole sequence of operations?

Input

The first line of input contains two integer: n, m (1 ≤ n, m ≤ 105). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — initial value of array elements.

Each of the next m lines begins with a number type <image>. 

  * If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1. 
  * If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 109), which correspond the operation 2. 
  * If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109), which correspond the operation 3. 

Output

For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

Examples

Input

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


Output

8
5


Input

10 10
6 9 6 7 6 1 10 10 9 5
1 3 9
2 7 10 9
2 5 10 8
1 4 7
3 3 7
2 7 9 9
1 2 4
1 6 6
1 5 9
3 1 10


Output

49
15
23
1
9

Note

Consider the first testcase:

  * At first, a = {1, 2, 3, 4, 5}. 
  * After operation 1, a = {1, 2, 3, 0, 1}. 
  * After operation 2, a = {1, 2, 5, 0, 1}. 
  * At operation 3, 2 + 5 + 0 + 1 = 8. 
  * After operation 4, a = {1, 2, 2, 0, 1}. 
  * At operation 5, 1 + 2 + 2 = 5. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 5;
int n, m;
long long sum[N * 4];
int max_[N * 4], a[N];
inline void pull(int t) {
  sum[t] = sum[2 * t] + sum[2 * t + 1];
  max_[t] = max(max_[2 * t], max_[2 * t + 1]);
}
void build(int t, int L, int R) {
  if (L == R) {
    sum[t] = a[L];
    max_[t] = a[L];
  } else {
    int M = (L + R) / 2;
    build(2 * t, L, M);
    build(2 * t + 1, M + 1, R);
    pull(t);
  }
}
long long q_sum(int t, int L, int R, int x, int y) {
  if (y < L || R < x)
    return 0;
  else if (x <= L && R <= y)
    return sum[t];
  else {
    int M = (L + R) / 2;
    return q_sum(2 * t, L, M, x, y) + q_sum(2 * t + 1, M + 1, R, x, y);
  }
}
void mod(int t, int L, int R, int x, int y, int v) {
  if (y < L || R < x) return;
  if (max_[t] < v) return;
  if (L == R) {
    sum[t] = sum[t] % v;
    max_[t] = max_[t] % v;
    return;
  }
  int M = (L + R) / 2;
  mod(2 * t, L, M, x, y, v);
  mod(2 * t + 1, M + 1, R, x, y, v);
  pull(t);
}
void change(int t, int L, int R, int x, int v) {
  if (L == R) {
    sum[t] = v;
    max_[t] = v;
  } else {
    int M = (L + R) / 2;
    if (x <= M)
      change(2 * t, L, M, x, v);
    else
      change(2 * t + 1, M + 1, R, x, v);
    pull(t);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  build(1, 1, n);
  int op, b, c, d;
  while (m--) {
    scanf("%d%d%d", &op, &b, &c);
    if (op == 1) {
      printf("%I64d\n", q_sum(1, 1, n, b, c));
    } else if (op == 2) {
      scanf("%d", &d);
      mod(1, 1, n, b, c, d);
    } else {
      change(1, 1, n, b, c);
    }
  }
  return 0;
}