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 NUM = 20;
int n, m;
set<int> p[NUM];
long long ans = 0;
long long get(long long a) { return a * (a + 1) / 2; }
void ins(int i, int a) {
  int j;
  for ((j) = (0); (j) != (NUM); ++(j)) {
    set<int>::iterator it = p[j].lower_bound(i);
    if (a & (1 << j) && *it == i) {
      int r = *(++it);
      int l = *(--(--it));
      ans += (get(r - l - 1) - get(i - l - 1) - get(r - i - 1)) << j;
      p[j].erase(i);
    }
    if (!(a & (1 << j)) && *it != i) {
      int r = *it;
      int l = *(--it);
      ans += (-get(r - l - 1) + get(i - l - 1) + get(r - i - 1)) << j;
      p[j].insert(i);
    }
  }
}
int main() {
  int i, j;
  scanf("%d%d", &n, &m);
  for ((j) = (0); (j) != (NUM); ++(j))
    for ((i) = (0); (i) <= (n + 1); ++(i)) p[j].insert(i);
  for ((i) = (1); (i) <= (n); ++(i)) {
    int a;
    scanf("%d", &a);
    ins(i, a);
  }
  for ((i) = (1); (i) <= (m); ++(i)) {
    int p, v;
    scanf("%d%d", &p, &v);
    ins(p, v);
    cout << ans << endl;
  }
  return 0;
}