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;
template <typename T, typename L>
inline bool smax(T &x, L y) {
  return x < y ? (x = y, true) : false;
}
template <typename T, typename L>
inline bool smin(T &x, L y) {
  return y < x ? (x = y, true) : false;
}
inline void build(int node, int b, int e);
inline long long query(int node, int b, int e, int l, int r);
inline void update(int node, int b, int e, int l, int r, int val);
inline void modify(int node, int b, int e, int l, int r, int val);
constexpr int MAXN = 1e5 + 15;
int n, q;
pair<long long, long long> tree[MAXN << 2];
long long a[MAXN];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  build(1, 0, n);
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int l, r;
      cin >> l >> r;
      cout << query(1, 0, n, l - 1, r) << '\n';
    } else if (type == 2) {
      int l, r, x;
      cin >> l >> r >> x;
      update(1, 0, n, l - 1, r, x);
    } else {
      int l, x;
      cin >> l >> x;
      modify(1, 0, n, l - 1, l, x);
    }
  }
  return false;
}
inline void build(int node, int b, int e) {
  if (b == e - 1) {
    tree[node] = make_pair(a[b], a[b]);
    return;
  }
  int mid = (b + e) >> 1;
  build(node << 1, b, mid);
  build(node << 1 | 1, mid, e);
  tree[node].first = tree[node << 1].first + tree[node << 1 | 1].first;
  tree[node].second = max(tree[node << 1].second, tree[node << 1 | 1].second);
}
inline long long query(int node, int b, int e, int l, int r) {
  if (b >= r || e <= l) {
    return 0;
  }
  if (b >= l && e <= r) {
    return tree[node].first;
  }
  int mid = (b + e) >> 1;
  return query(node << 1, b, mid, l, r) + query(node << 1 | 1, mid, e, l, r);
}
inline void update(int node, int b, int e, int l, int r, int val) {
  if (b >= r || e <= l || tree[node].second < val) {
    return;
  }
  if (b == e - 1) {
    tree[node].first %= val;
    tree[node].second = tree[node].first;
    return;
  }
  int mid = (b + e) >> 1;
  update(node << 1, b, mid, l, r, val);
  update(node << 1 | 1, mid, e, l, r, val);
  tree[node].first = tree[node << 1].first + tree[node << 1 | 1].first;
  tree[node].second = max(tree[node << 1].second, tree[node << 1 | 1].second);
}
inline void modify(int node, int b, int e, int l, int r, int val) {
  if (b >= r || e <= l) {
    return;
  }
  if (b >= l && e <= r) {
    tree[node].first = tree[node].second = val;
    return;
  }
  int mid = (b + e) >> 1;
  modify(node << 1, b, mid, l, r, val);
  modify(node << 1 | 1, mid, e, l, r, val);
  tree[node].first = tree[node << 1].first + tree[node << 1 | 1].first;
  tree[node].second = max(tree[node << 1].second, tree[node << 1 | 1].second);
}