Problem: Inna is fed up with jokes about female logic. So she started using binary logic instead.

Inna has an array of n elements a1[1], a1[2], ..., a1[n]. Girl likes to train in her binary logic, so she does an exercise consisting of n stages: on the first stage Inna writes out all numbers from array a1, on the i-th (i ≥ 2) stage girl writes all elements of array ai, which consists of n - i + 1 integers; the k-th integer of array ai is defined as follows: ai[k] = ai - 1[k] AND ai - 1[k + 1]. Here AND is bit-wise binary logical operation.

Dima decided to check Inna's skill. He asks Inna to change array, perform the exercise and say the sum of all <image> elements she wrote out during the current exercise.

Help Inna to answer the questions!

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — size of array a1 and number of Dima's questions. Next line contains n integers a1[1], a1[2], ..., a1[n] (0 ≤ ai ≤ 105) — initial array elements.

Each of next m lines contains two integers — Dima's question description. Each question consists of two integers pi, vi (1 ≤ pi ≤ n; 0 ≤ vi ≤ 105). For this question Inna should make a1[pi] equals vi, and then perform the exercise. Please, note that changes are saved from question to question.

Output

For each question print Inna's answer on a single line.

Examples

Input

3 4
1 1 1
1 1
2 2
3 2
1 2


Output

6
4
7
12

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 100010;
int n, m;
int a[maxn];
struct SegTree {
  long long ret[maxn << 2];
  int pre[maxn << 2], suf[maxn << 2];
  void up(int l, int r, int m, int rt) {
    pre[rt] = pre[rt << 1];
    suf[rt] = suf[rt << 1 | 1];
    if (pre[rt << 1] == (m - l + 1)) pre[rt] += pre[rt << 1 | 1];
    if (suf[rt << 1 | 1] == (r - m)) suf[rt] += suf[rt << 1];
    ret[rt] = ret[rt << 1] + ret[rt << 1 | 1];
    if (suf[rt << 1] && pre[rt << 1 | 1])
      ret[rt] += 1LL * suf[rt << 1] * pre[rt << 1 | 1];
  }
  void build(int l, int r, int rt, int p) {
    if (l == r) {
      ret[rt] = pre[rt] = suf[rt] = (a[l] & (1 << p)) == 0 ? 0 : 1;
      return;
    }
    int m = (l + r) >> 1;
    build(l, m, rt << 1, p);
    build(m + 1, r, rt << 1 | 1, p);
    up(l, r, m, rt);
  }
  void update(int x, int val, int l, int r, int rt) {
    if (l == r) {
      ret[rt] = pre[rt] = suf[rt] = val;
      return;
    }
    int m = (l + r) >> 1;
    if (x <= m)
      update(x, val, l, m, rt << 1);
    else
      update(x, val, m + 1, r, rt << 1 | 1);
    up(l, r, m, rt);
  }
} sg[20];
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int p = 0; p <= 18; p++) {
    sg[p].build(1, n, 1, p);
  }
  while (m--) {
    int id, v;
    long long ans = 0;
    scanf("%d%d", &id, &v);
    for (int p = 0; p <= 18; p++) {
      sg[p].update(id, (v & (1 << p)) == 0 ? 0 : 1, 1, n, 1);
      ans += 1LL * sg[p].ret[1] * (1 << p);
    }
    cout << ans << endl;
  }
}