Educational Codeforces Round 65 (Rated for Div. 2)E. Range Deleting【二分】

E. Range Deleting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array consisting of nn integers a1,a2,…,ana1,a2,…,an and an integer xx. It is guaranteed that for every ii, 1≤ai≤x1≤ai≤x.

Let's denote a function f(l,r)f(l,r) which erases all values such that l≤ai≤rl≤ai≤r from the array aa and returns the resulting array. For example, if a=[4,1,1,4,5,2,4,3]a=[4,1,1,4,5,2,4,3], then f(2,4)=[1,1,5]f(2,4)=[1,1,5].

Your task is to calculate the number of pairs (l,r)(l,r) such that 1≤l≤r≤x1≤l≤r≤x and f(l,r)f(l,r) is sorted in non-descending order. Note that the empty array is also considered sorted.

Input

The first line contains two integers nn and xx (1≤n,x≤1061≤n,x≤106) — the length of array aa and the upper limit for its elements, respectively.

The second line contains nn integers a1,a2,…ana1,a2,…an (1≤ai≤x1≤ai≤x).

Output

Print the number of pairs 1≤l≤r≤x1≤l≤r≤x such that f(l,r)f(l,r) is sorted in non-descending order.

Examples

input

Copy

3 3
2 3 1

output

Copy

4

input

Copy

7 4
1 3 1 2 2 4 3

output

Copy

6

Note

In the first test case correct pairs are (1,1)(1,1), (1,2)(1,2), (1,3)(1,3) and (2,3)(2,3).

In the second test case correct pairs are (1,3)(1,3), (1,4)(1,4), (2,3)(2,3), (2,4)(2,4), (3,3)(3,3) and (3,4)(3,4).

 

 

分析：不降序列要求满足，a[i].w

因为要求删除的是w连续的区间，所以直接对原数组重新排序，即问题变成了有多少个连续的区间使得删去这个区间后剩下的数组是单调的。类似于HihoCoder - 1797，不过一个是计数，一个是求最大值，大同小异，找左边最长连续，右边最长连续，然后二分一下计数即可。

#include "bits/stdc++.h"

using namespace std;

struct node {
    long long w, id;

    bool friend operator<(node a, node b) {
        if (a.w == b.w)return a.id < b.id;
        else return a.w < b.w;
    }
} a[1000004];

long long b[1000004][4];
long long l[1000004];
long long r[1000004];

int main() {
    long long n, x;
    cin >> n >> x;
    for (int i = 1; i <= n; ++i) {
        scanf("%lld", &a[i].w);
        a[i].id = i;
    }
    sort(a + 1, a + 1 + n);
    long long cnt = 1;
    b[cnt][0] = b[cnt][1] = a[1].id;
    b[cnt][3] = a[1].w;
    for (int i = 2; i <= n; ++i) {
        if (a[i].w == a[i - 1].w)b[cnt][1] = a[i].id;
        else {
            cnt++;
            b[cnt][0] = b[cnt][1] = a[i].id;
            b[cnt][3] = a[i].w;
        }
    }
    long long ans = 0;
    long long base = 0;
    long long lmax = 1;
    long long rmin = cnt;
    for (int i = 1; i <= cnt; ++i) {
        if (b[i][0] > base) {
            base = b[i][1];
            l[i] = base;
            lmax = i;
        } else break;
    }
    base = 1e9;
    for (int i = cnt; i >= 1; --i) {
        if (b[i][1] < base) {
            base = b[i][0];
            r[i] = base;
            rmin = i;
        } else break;
    }
    if (lmax < rmin) {
        for (int i = 1; i <= lmax; ++i) {
            int pos = upper_bound(r + (int) rmin, r + (int) cnt + 1, l[i]) - r;
            if (pos < cnt + 1)
                ans += (b[cnt][3] - b[pos - 1][3]) * (b[i + 1][3] - b[i][3]);
        }
        cout << ans + (b[lmax + 1][3] - b[1][3]) * (x - b[cnt][3] + 1) + (b[cnt][3] - b[rmin - 1][3]) * (b[1][3]) +
                (x - b[cnt][3] + 1) * (b[1][3]) << endl;
    } else {
        cout << x * (x + 1) / 2 << endl;
    }

}