so easy（并查集+unordered_map）

There are nn points in an array with index from 11 to nn, and there are two operations to those points.

1: 1 \ x1 x marking the point xx is not available

2: 2 \ x2 x query for the index of the first available point after that point (including xx itself) .

Input

n\quad qnq

z_1 \quad x_1z1​x1​

\vdots⋮

z_q\quad x_qzq​xq​

qq is the number of queries, zz is the type of operations, and xx is the index of operations. 1≤x<n<10^91≤x<n<109， 1 \leq q<10^61≤q<106 and zz is 11 or 22

Output

Output the answer for each query.

样例输入复制

5 3
1 2
2 2
2 1

样例输出复制

3
1

#include <bits/stdc++.h>

using namespace std;
const int maxn = 1e5 + 100;
unordered_map<int, int> fa;

int findfa(int x) {
    if (!fa.count(x)) return x;
    return fa[x] = findfa(fa[x]);
}


int main() {
    int n, q;
    scanf("%d %d", &n, &q);
    int op, x;
    while (q--) {
        scanf("%d %d", &op, &x);
        if (op == 1) {
            fa[x] = findfa(x + 1);
        } else {
            int ans = findfa(x);
            if (ans > n) ans = -1;
            printf("%d\n", ans);
        }
    }
    return 0;
}