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;
int a[100010][25], n, m, x, y;
long long s[30], f[100100];
struct bit {
  int t[100100];
  void add(int x, int y) {
    for (; x <= 100000; x += x & (-x)) t[x] += y;
  }
  int sum(int x) {
    int s = 0;
    for (; x; x -= x & (-x)) s += t[x];
    return s;
  }
  int ask(int l, int r) { return sum(r) - sum(l - 1); }
} t[30];
void init() {
  scanf("%d%d", &n, &m);
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &x);
    for (int j = (0); j <= (20); ++j) {
      a[i][j] = x & 1;
      if (x & 1) t[j].add(i, 1);
      x >>= 1;
    }
  }
  for (int i = (1); i <= (n); ++i) f[i] = (long long)i * (i + 1) / 2;
  for (int i = (0); i <= (20); ++i) s[i] = 0;
  for (int j = (0); j <= (20); ++j) {
    x = 0;
    for (int i = (1); i <= (n + 1); ++i)
      if (a[i][j])
        x++;
      else {
        s[j] += f[x];
        x = 0;
      }
  }
}
int find(int y, int x) {
  int l = 1, r = y, ll, k;
  if (y < 1 || y > n) return 0;
  if (!t[x].ask(y, y)) return 0;
  while (l < r) {
    k = (l + r) / 2;
    if (t[x].ask(k, y) == y - k + 1)
      r = k;
    else
      l = k + 1;
  }
  ll = l;
  l = y;
  r = n;
  while (l < r) {
    k = (l + r + 1) / 2;
    if (t[x].ask(y, k) == k - y + 1)
      l = k;
    else
      r = k - 1;
  }
  return r - ll + 1;
}
void deal(int y, int x) {
  if (!a[y][x]) {
    s[x] -= f[find(y, x)];
    t[x].add(y, -1);
    s[x] += f[find(y - 1, x)] + f[find(y + 1, x)];
  } else {
    s[x] -= f[find(y - 1, x)] + f[find(y + 1, x)];
    t[x].add(y, 1);
    s[x] += f[find(y, x)];
  }
}
void work() {
  while (m--) {
    long long sum = 0;
    scanf("%d%d", &y, &x);
    for (int i = (0); i <= (20); ++i) {
      if ((x & 1) != a[y][i]) {
        a[y][i] = x & 1;
        deal(y, i);
      }
      x >>= 1;
    }
    for (int i = (0); i <= (20); ++i) sum += ((long long)1 << i) * s[i];
    printf("%I64d\n", sum);
  }
}
int main() {
  init();
  work();
  return 0;
}