#include <bits/stdc++.h>
using namespace std;
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define loop(i, a) for (int i = 0; i < (a); ++i)
#define cont(i, a) for (int i = 1; i <= (a); ++i)
#define circ(i, a, b) for (int i = (a); i <= (b); ++i)
#define range(i, a, b, c) for (int i = (a); (c) > 0 ? i <= (b) : i >= (b); i += (c))
#define parse(it, x) for (auto &it : (x))
#define pub push_back
#define pob pop_back
#define emb emplace_back
#define mak make_pair
#define mkt make_tuple
typedef long long ll;
typedef long double lf;
const int Inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fll;

const int maxn = 25005, maxq = 100005;
int n, q;
int a[maxn], le[maxn];
pair<int, int> pp[maxn], ps[maxn], sp[maxn], ss[maxn];

int main() {
    scanf("%d%d", &n, &q);
    cont(i, n) scanf("%d", a + i);
    cont(i, n) {
        pp[i] = mak(0, 0);
        ps[i] = mak(n + 1, 0);
        sp[i] = mak(0, 0);
        ss[i] = mak(n + 1, 0);
        le[i] = i;
    }
    int ans = 0;
    while (q--) {
        int l, r; scanf("%d%d", &l, &r);
        int now = r + 1;
        while (le[now - 1] > l) --now;
        for (; now <= r; ++now) {
            while (le[now] > l) {
                --le[now];
                int pre = le[now];
                if (a[pre] < a[now] && a[now] < ss[pre].first) {
                    if (ss[pre].second) {
                        int nxt = ss[pre].second;
                        pp[nxt].second = 0;
                        --ans;
                    }
                    ss[pre] = mak(a[now], now);
                    if (pp[now].second) {
                        int nxt = pp[now].second;
                        ss[nxt].second = 0;
                        --ans;
                    }
                    pp[now] = mak(a[pre], pre);
                    ++ans;
                }
                if (a[pre] > a[now] && a[now] > sp[pre].first) {
                    if (sp[pre].second) {
                        int nxt = sp[pre].second;
                        ps[nxt].second = 0;
                        --ans;
                    }
                    sp[pre] = mak(a[now], now);
                    if (ps[now].second) {
                        int nxt = ps[now].second;
                        sp[nxt].second = 0;
                        --ans;
                    }
                    ps[now] = mak(a[pre], pre);
                    ++ans;
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

### Prompt

In cpp, your task is to solve the following 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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define loop(i, a) for (int i = 0; i < (a); ++i)
#define cont(i, a) for (int i = 1; i <= (a); ++i)
#define circ(i, a, b) for (int i = (a); i <= (b); ++i)
#define range(i, a, b, c) for (int i = (a); (c) > 0 ? i <= (b) : i >= (b); i += (c))
#define parse(it, x) for (auto &it : (x))
#define pub push_back
#define pob pop_back
#define emb emplace_back
#define mak make_pair
#define mkt make_tuple
typedef long long ll;
typedef long double lf;
const int Inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fll;

const int maxn = 25005, maxq = 100005;
int n, q;
int a[maxn], le[maxn];
pair<int, int> pp[maxn], ps[maxn], sp[maxn], ss[maxn];

int main() {
    scanf("%d%d", &n, &q);
    cont(i, n) scanf("%d", a + i);
    cont(i, n) {
        pp[i] = mak(0, 0);
        ps[i] = mak(n + 1, 0);
        sp[i] = mak(0, 0);
        ss[i] = mak(n + 1, 0);
        le[i] = i;
    }
    int ans = 0;
    while (q--) {
        int l, r; scanf("%d%d", &l, &r);
        int now = r + 1;
        while (le[now - 1] > l) --now;
        for (; now <= r; ++now) {
            while (le[now] > l) {
                --le[now];
                int pre = le[now];
                if (a[pre] < a[now] && a[now] < ss[pre].first) {
                    if (ss[pre].second) {
                        int nxt = ss[pre].second;
                        pp[nxt].second = 0;
                        --ans;
                    }
                    ss[pre] = mak(a[now], now);
                    if (pp[now].second) {
                        int nxt = pp[now].second;
                        ss[nxt].second = 0;
                        --ans;
                    }
                    pp[now] = mak(a[pre], pre);
                    ++ans;
                }
                if (a[pre] > a[now] && a[now] > sp[pre].first) {
                    if (sp[pre].second) {
                        int nxt = sp[pre].second;
                        ps[nxt].second = 0;
                        --ans;
                    }
                    sp[pre] = mak(a[now], now);
                    if (ps[now].second) {
                        int nxt = ps[now].second;
                        sp[nxt].second = 0;
                        --ans;
                    }
                    ps[now] = mak(a[pre], pre);
                    ++ans;
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

```