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 = 100050;
int n, q;
struct Node {
  long long sum, maxVal;
  int from;
} node[N << 2];
inline void pushUp(int rt) {
  node[rt].sum = node[rt << 1].sum + node[rt << 1 | 1].sum;
  if (node[rt << 1].maxVal < node[rt << 1 | 1].maxVal) {
    node[rt].maxVal = node[rt << 1 | 1].maxVal;
    node[rt].from = node[rt << 1 | 1].from;
  } else {
    node[rt].maxVal = node[rt << 1].maxVal;
    node[rt].from = node[rt << 1].from;
  }
}
inline void build(int l, int r, int rt) {
  if (l == r) {
    scanf("%I64d", &node[rt].sum);
    node[rt].maxVal = node[rt].sum;
    node[rt].from = l;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  pushUp(rt);
}
inline void assign(int p, int val, int l, int r, int rt) {
  if (l == r) {
    node[rt].sum = node[rt].maxVal = val;
    return;
  }
  int m = (l + r) >> 1;
  if (p <= m)
    assign(p, val, l, m, rt << 1);
  else
    assign(p, val, m + 1, r, rt << 1 | 1);
  pushUp(rt);
}
inline pair<int, int> queryMax(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) return make_pair(node[rt].maxVal, node[rt].from);
  int m = (l + r) >> 1;
  pair<int, int> ret = make_pair(-1, 1);
  if (L <= m) ret = queryMax(L, R, l, m, rt << 1);
  if (R > m) ret = max(ret, queryMax(L, R, m + 1, r, rt << 1 | 1));
  return ret;
}
inline long long query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) {
    return node[rt].sum;
  }
  int m = (l + r) >> 1;
  long long ret = 0;
  if (L <= m) ret += query(L, R, l, m, rt << 1);
  if (R > m) ret += query(L, R, m + 1, r, rt << 1 | 1);
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  build(1, n, 1);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%I64d\n", query(l, r, 1, n, 1));
    } else if (op == 2) {
      int l, r, x;
      scanf("%d%d%d", &l, &r, &x);
      while (true) {
        pair<int, int> ret = queryMax(l, r, 1, n, 1);
        if (ret.first >= x) {
          assign(ret.second, ret.first % x, 1, n, 1);
        } else
          break;
      }
    } else {
      int k, x;
      scanf("%d%d", &k, &x);
      assign(k, x, 1, n, 1);
    }
  }
  return 0;
}