You are given a set of $\mathrm{nn segments\; on\; the\; axis OxOx,\; each\; segment\; has\; integer\; endpoints\; between 11 and mm inclusive.\; Segments\; may\; intersect,\; overlap\; or\; even\; coincide\; with\; each\; other.\; Each\; segment\; is\; characterized\; by\; two\; integers lili and riri (1\le li\le ri\le m1\le li\le ri\le m)\; —\; coordinates\; of\; the\; left\; and\; of\; the\; right\; endpoints.}$

Consider all integer points between $\mathrm{11 and mm inclusive.\; Your\; task\; is\; to\; print\; all\; such\; points\; that\; don\text{'}t\; belong\; to\; any\; segment.\; The\; point xxbelongs\; to\; the\; segment [l;r][l;r] if\; and\; only\; if l\le x\le rl\le x\le r.}$

Input

The first line of the input contains two integers $\mathrm{nn and mm (1\le n,m\le 1001\le n,m\le 100)\; —\; the\; number\; of\; segments\; and\; the\; upper\; bound\; for\; coordinates.}$

The next $\mathrm{nn lines\; contain\; two\; integers\; each lili and riri (1\le li\le ri\le m1\le li\le ri\le m)\; —\; the\; endpoints\; of\; the ii-th\; segment.\; Segments\; may\; intersect,\; overlap\; or\; even\; coincide\; with\; each\; other.\; Note,\; it\; is\; possible\; that li=rili=ri,\; i.e.\; a\; segment\; can\; degenerate\; to\; a\; point.}$

Output

In the first line print one integer $\mathrm{kk —\; the\; number\; of\; points\; that\; don\text{'}t\; belong\; to\; any\; segment.}$

In the second line print exactly $\mathrm{kk integers\; in any order\; —\; the\; points\; that\; don\text{'}t\; belong\; to\; any\; segment.\; All\; points\; you\; print\; should\; be\; distinct.}$

If there are no such points at all, print a single integer $\mathrm{00 in\; the\; first\; line\; and\; either\; leave\; the\; second\; line\; empty\; or\; do\; not\; print\; it\; at\; all.}$

Examples

input

Copy

3 5 2 2 1 2 5 5

output

Copy

2 3 4

input

Copy

1 7 1 7

output

Copy

0

代码：

#include 
     
      using namespace std;int N, M;int vis[110], num[110];int main() {    memset(vis, 0, sizeof(vis));    scanf("%d%d", &N, &M);    for(int i = 1; i <= N; i ++) {        int l, r;        scanf("%d%d", &l, &r);        for(int j = l; j <= r; j ++)            vis[j] = 1;    }    int ans = 0;    for(int i = 1; i <= M; i ++) {        if(!vis[i]) {            ans ++;            num[ans] = i;        }    }    printf("%d\n", ans);    for(int i = 1; i <= ans; i ++) {        printf("%d", num[i]);        printf("%s", i != ans ? " " : "\n");    }    return 0;}