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 long long maxn = 100005;
long long n, m;
long long c[maxn];
struct sg_tr {
  long long l, r, sum, mx;
};
sg_tr a[maxn << 2];
inline long long grd() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
inline void upd(long long p) {
  a[p].sum = a[(p << 1)].sum + a[((p << 1) | 1)].sum;
  a[p].mx = std::max(a[(p << 1)].mx, a[((p << 1) | 1)].mx);
}
void bld(long long p, long long l, long long r) {
  a[p].l = l, a[p].r = r;
  if (l == r) {
    a[p].sum = a[p].mx = c[l];
    return;
  }
  long long mid = (l + r) >> 1;
  bld((p << 1), l, mid);
  bld(((p << 1) | 1), mid + 1, r);
  upd(p);
}
void mdf(long long p, long long ql, long long qr, long long mod) {
  if (a[p].mx < mod) return;
  if (a[p].l == a[p].r) {
    a[p].sum %= mod;
    a[p].mx %= mod;
    return;
  }
  long long mid = (a[p].l + a[p].r) >> 1;
  if (qr <= mid)
    mdf((p << 1), ql, qr, mod);
  else if (ql > mid)
    mdf(((p << 1) | 1), ql, qr, mod);
  else
    mdf((p << 1), ql, mid, mod), mdf(((p << 1) | 1), mid + 1, qr, mod);
  upd(p);
}
void replc(long long p, long long pos, long long val) {
  if (a[p].l == a[p].r) {
    a[p].sum = a[p].mx = val;
    return;
  }
  long long mid = (a[p].l + a[p].r) >> 1;
  if (pos <= mid)
    replc((p << 1), pos, val);
  else
    replc(((p << 1) | 1), pos, val);
  upd(p);
}
long long qry(long long p, long long ql, long long qr) {
  if (a[p].l == ql && a[p].r == qr) return a[p].sum;
  long long mid = (a[p].l + a[p].r) >> 1;
  if (qr <= mid)
    return qry((p << 1), ql, qr);
  else if (ql > mid)
    return qry(((p << 1) | 1), ql, qr);
  return qry((p << 1), ql, mid) + qry(((p << 1) | 1), mid + 1, qr);
}
int main() {
  n = grd(), m = grd();
  for (register long long i = 1; i <= n; i++) c[i] = grd();
  bld(1, 1, n);
  while (m--) {
    long long opt = grd(), x = grd(), y = grd();
    switch (opt) {
      case 1:
        printf("%lld\n", qry(1, x, y));
        break;
      case 3:
        replc(1, x, y);
        break;
      case 2:
        long long z = grd();
        mdf(1, x, y, z);
        break;
    }
  }
  return 0;
}