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 = 200005;
long long sum[maxn << 2], maxv[maxn << 2], a[maxn << 2];
long long n, m, ans;
inline long long read() {
  long long k = 0, f = 1;
  char ch = getchar();
  while (ch < 48 || ch > 57) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= 48 && ch <= 57) {
    k = (k << 1) + (k << 3) + (ch ^ 48);
    ch = getchar();
  }
  return k * f;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
void pushup(long long u) {
  sum[u] = sum[(u << 1)] + sum[(u << 1) | 1];
  maxv[u] = max(maxv[(u << 1)], maxv[(u << 1) | 1]);
}
void update(long long rt, long long l, long long r, long long u, long long v) {
  if (r < u || l > u) return;
  if (r == l && u == r) {
    sum[rt] = v;
    maxv[rt] = v;
    return;
  }
  long long mid = (l + r) >> 1;
  update((rt << 1), l, mid, u, v);
  update((rt << 1) | 1, mid + 1, r, u, v);
  pushup(rt);
  return;
}
void update1(long long u, long long l, long long r, long long ll, long long rr,
             long long v) {
  if (ll > r || rr < l) return;
  if (maxv[u] < v) return;
  if (l == r) {
    maxv[u] %= v;
    sum[u] %= v;
    return;
  }
  long long mid = (l + r) >> 1;
  update1((u << 1), l, mid, ll, rr, v);
  update1((u << 1) | 1, mid + 1, r, ll, rr, v);
  pushup(u);
}
void build(long long u, long long l, long long r) {
  if (l == r) {
    sum[u] = a[l];
    maxv[u] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build((u << 1), l, mid);
  build((u << 1) | 1, mid + 1, r);
  pushup(u);
}
void query(long long u, long long l, long long r, long long ll, long long rr) {
  if (l > rr || r < ll) return;
  if (ll <= l && r <= rr) {
    ans += sum[u];
    return;
  }
  long long mid = (l + r) >> 1;
  query((u << 1), l, mid, ll, rr);
  query((u << 1) | 1, mid + 1, r, ll, rr);
  return;
}
signed main() {
  n = read();
  m = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  build(1, 1, n);
  while (m--) {
    long long op, l, r, t;
    op = read();
    l = read();
    r = read();
    if (op == 3) {
      update(1, 1, n, l, r);
    }
    if (op == 2) t = read(), update1(1, 1, n, l, r, t);
    if (op == 1) {
      ans = 0;
      query(1, 1, n, l, r);
      printf("%lld\n", ans);
    }
  }
}