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 N = 400100;
int a[N];
struct SegTree {
  long long sum[N], pre[N], suf[N], tp;
  void PushUp(int l, int r, int rt) {
    int L = rt << 1, R = rt << 1 | 1;
    int m = (l + r) >> 1;
    sum[rt] = sum[L] + sum[R];
    pre[rt] = pre[L];
    suf[rt] = suf[R];
    if (pre[L] == m - l + 1) pre[rt] = pre[L] + pre[R];
    if (suf[R] == r - m) suf[rt] = suf[L] + suf[R];
    if (suf[L] && pre[R]) {
      sum[rt] -= suf[L] * (suf[L] + 1) / 2 + pre[R] * (pre[R] + 1) / 2;
      long long sm = suf[L] + pre[R];
      sum[rt] += sm * (sm + 1) / 2;
    }
  }
  void build(int l, int r, int rt) {
    if (l == r) {
      pre[rt] = suf[rt] = sum[rt] = ((a[l] >> tp) & 1);
      return;
    }
    int m = (l + r) >> 1;
    build(l, m, rt << 1);
    build(m + 1, r, rt << 1 | 1);
    PushUp(l, r, rt);
  }
  void update(int p, int sc, int l, int r, int rt) {
    if (l == r) {
      sum[rt] = pre[rt] = suf[rt] = ((sc >> tp) & 1);
      return;
    }
    int m = (l + r) >> 1;
    if (p <= m)
      update(p, sc, l, m, rt << 1);
    else
      update(p, sc, m + 1, r, rt << 1 | 1);
    PushUp(l, r, rt);
  }
} st[20];
int main() {
  int n, m, p, v;
  while (scanf("%d%d", &n, &m) != EOF) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    long long sss = 0;
    for (int i = 0; i < 20; i++) st[i].tp = i, st[i].build(1, n, 1);
    while (m--) {
      scanf("%d%d", &p, &v);
      long long res = 0;
      for (int i = 0; i < 20; i++)
        st[i].update(p, v, 1, n, 1), res += st[i].sum[1] * (1 << i);
      printf("%I64d\n", res);
    }
  }
}