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 comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast", "unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long inf = numeric_limits<long long>::max();
const long double eps = 1e-9;
const long double pi = acos(-1);
inline void solve(), read();
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read();
  solve();
  return 0;
}
struct vertex {
  long long sum = 0, max = -inf, min = inf, push = -1;
};
struct segment {
  long long n;
  vector<vertex> t;
  segment() {}
  segment(vector<long long>& a) {
    long long l = log2((long long)((a).size())) + 1;
    n = 1ll << l;
    t.resize(4 * n);
    build(a);
  }
  vertex merge(vertex v1, vertex v2) {
    vertex v;
    v.sum = v1.sum + v2.sum;
    v.max = max(v1.max, v2.max);
    v.min = min(v1.min, v2.min);
    v.push = -1;
    return v;
  }
  void build(vector<long long>& a) {
    for (long long i = 0; i < (long long)((a).size()); i++) {
      t[i + n].sum = t[i + n].max = t[i + n].min = a[i];
    }
    for (long long i = n - 1; i >= 0; i--) {
      t[i] = merge(t[2 * i], t[2 * i + 1]);
    }
  }
  void push(long long v, long long tl, long long tr) {
    if (t[v].push == -1) {
      return;
    }
    t[v].sum = t[v].push * (tr - tl + 1);
    t[2 * v].push = t[v].push;
    t[2 * v + 1].push = t[v].push;
    t[v].max = t[v].min = t[v].push;
    t[v].push = -1;
  }
  void modulo(long long v, long long tl, long long tr, long long l, long long r,
              long long x) {
    push(v, tl, tr);
    if (tl > r || tr < l || t[v].max < x) {
      return;
    }
    if (l <= tl && tr <= r && t[v].max == t[v].min) {
      t[v].push = t[v].max % x;
      push(v, tl, tr);
      return;
    }
    long long tm = (tl + tr) / 2;
    modulo(2 * v, tl, tm, l, r, x);
    modulo(2 * v + 1, tm + 1, tr, l, r, x);
    t[v] = merge(t[2 * v], t[2 * v + 1]);
  }
  void upd(long long v, long long tl, long long tr, long long pos,
           long long x) {
    push(v, tl, tr);
    if (tl > pos || tr < pos) {
      return;
    }
    if (tl == tr) {
      t[v].push = x;
      push(v, tl, tr);
      return;
    }
    long long tm = (tl + tr) / 2;
    upd(2 * v, tl, tm, pos, x);
    upd(2 * v + 1, tm + 1, tr, pos, x);
    t[v] = merge(t[2 * v], t[2 * v + 1]);
  }
  long long sum(long long v, long long tl, long long tr, long long l,
                long long r) {
    push(v, tl, tr);
    if (tl > r || tr < l) {
      return 0;
    }
    if (l <= tl && tr <= r) {
      return t[v].sum;
    }
    long long tm = (tl + tr) / 2;
    long long v1 = sum(2 * v, tl, tm, l, r);
    long long v2 = sum(2 * v + 1, tm + 1, tr, l, r);
    return v1 + v2;
  }
};
long long n, m;
vector<long long> a;
segment st;
inline void solve() {
  st = segment(a);
  for (long long i = 0; i < m; i++) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long l, r;
      cin >> l >> r;
      --l, --r;
      long long ans = st.sum(1, 0, st.n - 1, l, r);
      cout << ans << "\n";
    } else if (t == 2) {
      long long l, r, x;
      cin >> l >> r >> x;
      --l, --r;
      st.modulo(1, 0, st.n - 1, l, r, x);
    } else {
      long long k, x;
      cin >> k >> x;
      --k;
      st.upd(1, 0, st.n - 1, k, x);
    }
  }
}
inline void read() {
  cin >> n >> m;
  a.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
}