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>
#pragma GCC optimize("Ofast")
using namespace std;
const int N = 1e5 + 5;
int maxi[N << 2], a[N], n, m;
long long sum[N << 2];
void modify_s(int k, int l, int r, int pos, int val) {
  if (l == r) {
    maxi[k] = sum[k] = val;
    return;
  }
  if (pos <= ((l + r) >> 1))
    modify_s(k << 1, l, ((l + r) >> 1), pos, val);
  else
    modify_s(k << 1 | 1, ((l + r) >> 1) + 1, r, pos, val);
  maxi[k] = max(maxi[k << 1], maxi[k << 1 | 1]),
  sum[k] = sum[k << 1] + sum[k << 1 | 1];
}
void modify_m(int k, int l, int r, int ql, int qr, int val) {
  if (l == r) {
    maxi[k] %= val, sum[k] %= val;
    return;
  }
  if (ql <= ((l + r) >> 1) && maxi[k << 1] >= val)
    modify_m(k << 1, l, ((l + r) >> 1), ql, qr, val);
  if (qr > ((l + r) >> 1) && maxi[k << 1 | 1] >= val)
    modify_m(k << 1 | 1, ((l + r) >> 1) + 1, r, ql, qr, val);
  maxi[k] = max(maxi[k << 1], maxi[k << 1 | 1]),
  sum[k] = sum[k << 1] + sum[k << 1 | 1];
}
long long query(int k, int l, int r, int ql, int qr) {
  if (l >= ql && r <= qr) return sum[k];
  long long ans = 0;
  if (ql <= ((l + r) >> 1)) ans += query(k << 1, l, ((l + r) >> 1), ql, qr);
  if (qr > ((l + r) >> 1))
    ans += query(k << 1 | 1, ((l + r) >> 1) + 1, r, ql, qr);
  return ans;
}
void build(int k, int l, int r) {
  if (l == r) {
    maxi[k] = sum[k] = a[l];
    return;
  }
  build(k << 1, l, ((l + r) >> 1));
  build(k << 1 | 1, ((l + r) >> 1) + 1, r);
  maxi[k] = max(maxi[k << 1], maxi[k << 1 | 1]),
  sum[k] = sum[k << 1] + sum[k << 1 | 1];
}
char bufin[1 << 21], *p1, *p2;
inline char gc() {
  if (p1 == p2) {
    p1 = bufin, p2 = p1 + fread(bufin, 1, 1 << 21, stdin);
    if (p1 == p2) return EOF;
  }
  return *p1++;
}
inline void read(int &ret) {
  char ch = gc();
  ret = 0;
  while (ch < '0' || ch > '9') ch = gc();
  while (ch >= '0' && ch <= '9') ret = ret * 10 + ch - '0', ch = gc();
}
char bufout[1 << 21], buf[20];
int t, top;
inline void flush() {
  fwrite(bufout, 1, t, stdout);
  t = 0;
}
inline void write(long long ret) {
  if (t + 20 >= (1 << 21)) flush();
  top = 0;
  if (ret == 0) bufout[t++] = '0';
  while (ret) buf[top++] = ret % 10 + '0', ret /= 10;
  for (register int i = top - 1; i >= 0; i--) bufout[t++] = buf[i];
  bufout[t++] = '\n';
}
signed main() {
  read(n), read(m);
  for (register int i = 1; i <= n; i++) read(a[i]);
  build(1, 1, n);
  while (m--) {
    int type, a, b, c;
    read(type), read(a), read(b);
    switch (type) {
      case 1:
        write(query(1, 1, n, a, b));
        break;
      case 2:
        read(c), modify_m(1, 1, n, a, b, c);
        break;
      default:
        modify_s(1, 1, n, a, b);
    }
  }
  flush();
  return 0;
}