Problem: Touko's favorite sequence of numbers is a permutation a_1, a_2, ..., a_n of 1, 2, ..., n, and she wants some collection of permutations that are similar to her favorite permutation.

She has a collection of q intervals of the form [l_i, r_i] with 1 ≤ l_i ≤ r_i ≤ n. To create permutations that are similar to her favorite permutation, she coined the following definition:

  * A permutation b_1, b_2, ..., b_n allows an interval [l', r'] to holds its shape if for any pair of integers (x, y) such that l' ≤ x < y ≤ r', we have b_x < b_y if and only if a_x < a_y. 
  * A permutation b_1, b_2, ..., b_n is k-similar if b allows all intervals [l_i, r_i] for all 1 ≤ i ≤ k to hold their shapes. 



Yuu wants to figure out all k-similar permutations for Touko, but it turns out this is a very hard task; instead, Yuu will encode the set of all k-similar permutations with directed acylic graphs (DAG). Yuu also coined the following definitions for herself:

  * A permutation b_1, b_2, ..., b_n satisfies a DAG G' if for all edge u → v in G', we must have b_u < b_v. 
  * A k-encoding is a DAG G_k on the set of vertices 1, 2, ..., n such that a permutation b_1, b_2, ..., b_n satisfies G_k if and only if b is k-similar. 



Since Yuu is free today, she wants to figure out the minimum number of edges among all k-encodings for each k from 1 to q.

Input

The first line contains two integers n and q (1 ≤ n ≤ 25 000, 1 ≤ q ≤ 100 000).

The second line contains n integers a_1, a_2, ..., a_n which form a permutation of 1, 2, ..., n.

The i-th of the following q lines contains two integers l_i and r_i. (1 ≤ l_i ≤ r_i ≤ n).

Output

Print q lines. The k-th of them should contain a single integer — The minimum number of edges among all k-encodings.

Examples

Input


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


Output


2
4
3


Input


8 4
3 7 4 8 1 5 2 6
3 6
1 6
3 8
1 8


Output


3
5
9
7


Input


10 10
10 5 1 2 7 3 9 4 6 8
2 2
4 5
6 8
4 10
4 4
2 7
2 2
7 8
3 7
2 10


Output


0
1
3
6
6
9
9
9
9
8

Note

For the first test case:

  * All 1-similar permutations must allow the interval [1, 3] to hold its shape. Therefore, the set of all 1-similar permutations is \{[3, 4, 2, 1], [3, 4, 1, 2], [2, 4, 1, 3], [2, 3, 1, 4]\}. The optimal encoding of these permutations is <image>
  * All 2-similar permutations must allow the intervals [1, 3] and [2, 4] to hold their shapes. Therefore, the set of all 2-similar permutations is \{[3, 4, 1, 2], [2, 4, 1, 3]\}. The optimal encoding of these permutations is <image>
  * All 3-similar permutations must allow the intervals [1, 3], [2, 4], and [1, 4] to hold their shapes. Therefore, the set of all 3-similar permutations only includes [2, 4, 1, 3]. The optimal encoding of this permutation is <image>

Solution (in CPP):

Solution:
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;

const int maxn = 25010;
int n, q, ans, a[maxn], lb[maxn], rb[maxn];
int prel[maxn], prer[maxn], nxtl[maxn], nxtr[maxn];

int main() {
    scanf("%d %d", &n, &q);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]), lb[i] = rb[i] = i;
        prel[i] = prer[i] = nxtl[i] = nxtr[i] = -1;
    }
    while (q--) {
        int l, r; scanf("%d %d", &l, &r);
        for (int i = l; rb[i] < r; i++) {
            while (rb[i] < r) {
                int k = ++rb[i];
                if (a[k] < a[i]) {
                    if (~prer[i]) ans -= (nxtl[prer[i]] == i);
                    if (!~prer[i] || a[k] > a[prer[i]]) prer[i] = k;
                    if (~prer[i]) ans += (nxtl[prer[i]] == i);
                } else {
                    if (~nxtr[i]) ans -= (prel[nxtr[i]] == i);
                    if (!~nxtr[i] || a[k] < a[nxtr[i]]) nxtr[i] = k;
                    if (~nxtr[i]) ans += (prel[nxtr[i]] == i);
                }
            }
        }
        for (int i = r; lb[i] > l; i--) {
            while (lb[i] > l) {
                int k = --lb[i];
                if (a[k] < a[i]) {
                    if (~prel[i]) ans -= (nxtr[prel[i]] == i);
                    if (!~prel[i] || a[k] > a[prel[i]]) prel[i] = k;
                    if (~prel[i]) ans += (nxtr[prel[i]] == i);
                } else {
                    if (~nxtl[i]) ans -= (prer[nxtl[i]] == i);
                    if (!~nxtl[i] || a[k] < a[nxtl[i]]) nxtl[i] = k;
                    if (~nxtl[i]) ans += (prer[nxtl[i]] == i);
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}