#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
const int M = 1000005;
int go[M][4], fa[M], sum[M], val[M], lzy[M], hsc[M], tsz;
void pull(int x) {
  if (!x) return;
  sum[x] = val[x] + sum[go[x][0]] + sum[go[x][1]];
  if (go[x][2]) sum[x] += min(sum[go[x][2]], sum[go[x][3]]);
  hsc[x] = go[x][2] || hsc[go[x][0]] || hsc[go[x][1]];
}
int make(int v) {
  ++tsz;
  val[tsz] = v;
  pull(tsz);
  return tsz;
}
void rev(int x) {
  if (!x) return;
  swap(go[x][0], go[x][1]);
  lzy[x] ^= 1;
}
void push(int x) {
  if (lzy[x]) {
    rev(go[x][0]), rev(go[x][1]);
    if (go[x][2]) rev(go[x][2]), rev(go[x][3]);
    lzy[x] = 0;
  }
}
int is2(int x) { return go[fa[x]][0] == x || go[fa[x]][1] == x; }
int pd2(int x) { return go[fa[x]][0] == x ? 0 : go[fa[x]][1] == x ? 1 : -1; }
int is4(int x) { return go[fa[x]][2] == x || go[fa[x]][3] == x; }
int pd4(int x) { return go[fa[x]][2] == x ? 2 : go[fa[x]][3] == x ? 3 : -1; }
int isl(int x) { return !go[x][0] && !go[x][1]; }
void rot(int x) {
  int y = fa[x], z = fa[y], p = pd2(x), w = go[x][p ^ 1];
  if (is2(y))
    go[z][pd2(y)] = x;
  else if (is4(y))
    go[z][pd4(y)] = x;
  fa[x] = z;
  go[x][p ^ 1] = y;
  fa[y] = x;
  go[y][p] = w;
  if (w) fa[w] = y;
  pull(y);
}
void cl(int x) {
  if (pd2(x) != -1) cl(fa[x]);
  push(x);
}
void splay(int x, int y = 0) {
  for (cl(x); pd2(x) != -1 && fa[x] != y; rot(x)) {
    if (pd2(fa[x]) != -1 && fa[fa[x]] != y)
      rot(pd2(fa[x]) == pd2(x) ? fa[x] : x);
  }
  pull(x);
}
int fir(int x) {
  while (go[x][0]) push(x), x = go[x][0];
  return x;
}
void cyc(int c, int x) {
  int d = pd4(x);
  splay(c);
  push(x);
  int o = fir(go[c][1]);
  splay(o, c);
  go[c][1] = 0;
  go[c][d] = go[x][0];
  fa[go[c][d]] = c;
  go[x][0] = 0;
  rev(go[x][1]);
  go[o][1] = go[x][1];
  fa[go[o][1]] = o;
  go[x][1] = 0;
  go[o][0] = go[c][d ^ 1];
  fa[go[o][0]] = o;
  go[c][d ^ 1] = o;
  fa[o] = c;
  pull(o);
  pull(c);
}
void access(int x) {
  int y = 0, z = x;
  for (; x; y = x, x = fa[x]) {
    splay(x);
    if (fa[x] && go[fa[x]][2]) cyc(fa[x], x);
    go[x][1] = y;
    pull(x);
  }
  splay(z);
}
void mkrt(int x) {
  access(x);
  rev(x);
}
void path(int x, int y) {
  mkrt(x);
  access(y);
  splay(x, y);
}
int get(int x, int y) {
  if (x == y) return 0;
  path(x, y);
  if (fa[x] != y) return -1;
  return sum[y];
}
bool link(int x, int y, int w) {
  if (x == y) return false;
  path(x, y);
  if (fa[x] != y) {
    int e = make(w);
    fa[x] = e;
    fa[e] = y;
    return true;
  }
  if (hsc[y]) return false;
  int e = make(w), c = make(0);
  go[c][2] = go[x][1];
  fa[go[c][2]] = c;
  go[c][3] = e;
  fa[e] = c;
  go[x][1] = c;
  fa[c] = x;
  pull(c);
  pull(x);
  pull(y);
  return true;
}
bool cut(int x, int y, int w) {
  if (x == y) return false;
  path(x, y);
  if (fa[x] != y) return false;
  if (!isl(go[x][1])) return false;
  int c = go[x][1];
  push(c);
  if (go[c][2]) {
    int d = 0;
    if (isl(go[c][2]) && val[go[c][2]] == w) d = 2;
    if (isl(go[c][3]) && val[go[c][3]] == w) d = 3;
    if (!d) return false;
    go[x][1] = go[c][d ^ 1];
    fa[go[x][1]] = x;
    pull(x);
    pull(y);
    return true;
  } else {
    if (val[c] != w) return false;
    go[x][1] = 0;
    go[y][0] = 0;
    fa[x] = 0;
    pull(x);
    pull(y);
    return true;
  }
}
int n, m;
int U[500005], V[500005];
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  tsz = n;
  for (int i = 1; i <= m; i++) scanf("%d%d", &U[i], &V[i]);
  for (int i = 1, j = 0; i <= m; i++) {
    if (j < i) {
      j = i;
      link(U[j], V[j], 0);
    }
    while (j < m && link(U[j + 1], V[j + 1], 0)) j++;
    cut(U[i], V[i], 0);
    ans += j - i + 1;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

You are given an undirected graph of N nodes and M edges, E_1, E_2, ... E_M.

A connected graph is a cactus if each of it's edges belogs to at most one simple cycle. A graph is a desert if each of it's connected components is a cactus. 

Find the number of pairs (L, R), (1 ≤ L ≤ R ≤ M) such that, if we delete all the edges except for E_L, E_{L+1}, ... E_R, the graph is a desert. 

Input

The first line contains two integers N and M (2 ≤ N ≤ 2.5 × 10^5, 1 ≤ M ≤ 5 × 10^5). Each of the next M lines contains two integers. The i-th line describes the i-th edge. It contains integers U_i and V_i, the nodes connected by the i-th edge (E_i=(U_i, V_i)). It is guaranteed that 1 ≤ U_i, V_i ≤ N and U_i ≠ V_i. 

Output

The output contains one integer number – the answer.

Examples

Input


5 6
1 2
2 3
3 4
4 5
5 1
2 4


Output


20

Input


2 3
1 2
1 2
1 2


Output


5

Note

In the second example: Graphs for pairs (1, 1), (2, 2) and (3, 3) are deserts because they don't have any cycles. Graphs for pairs (1, 2) and (2, 3) have one cycle of length 2 so they are deserts.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
const int M = 1000005;
int go[M][4], fa[M], sum[M], val[M], lzy[M], hsc[M], tsz;
void pull(int x) {
  if (!x) return;
  sum[x] = val[x] + sum[go[x][0]] + sum[go[x][1]];
  if (go[x][2]) sum[x] += min(sum[go[x][2]], sum[go[x][3]]);
  hsc[x] = go[x][2] || hsc[go[x][0]] || hsc[go[x][1]];
}
int make(int v) {
  ++tsz;
  val[tsz] = v;
  pull(tsz);
  return tsz;
}
void rev(int x) {
  if (!x) return;
  swap(go[x][0], go[x][1]);
  lzy[x] ^= 1;
}
void push(int x) {
  if (lzy[x]) {
    rev(go[x][0]), rev(go[x][1]);
    if (go[x][2]) rev(go[x][2]), rev(go[x][3]);
    lzy[x] = 0;
  }
}
int is2(int x) { return go[fa[x]][0] == x || go[fa[x]][1] == x; }
int pd2(int x) { return go[fa[x]][0] == x ? 0 : go[fa[x]][1] == x ? 1 : -1; }
int is4(int x) { return go[fa[x]][2] == x || go[fa[x]][3] == x; }
int pd4(int x) { return go[fa[x]][2] == x ? 2 : go[fa[x]][3] == x ? 3 : -1; }
int isl(int x) { return !go[x][0] && !go[x][1]; }
void rot(int x) {
  int y = fa[x], z = fa[y], p = pd2(x), w = go[x][p ^ 1];
  if (is2(y))
    go[z][pd2(y)] = x;
  else if (is4(y))
    go[z][pd4(y)] = x;
  fa[x] = z;
  go[x][p ^ 1] = y;
  fa[y] = x;
  go[y][p] = w;
  if (w) fa[w] = y;
  pull(y);
}
void cl(int x) {
  if (pd2(x) != -1) cl(fa[x]);
  push(x);
}
void splay(int x, int y = 0) {
  for (cl(x); pd2(x) != -1 && fa[x] != y; rot(x)) {
    if (pd2(fa[x]) != -1 && fa[fa[x]] != y)
      rot(pd2(fa[x]) == pd2(x) ? fa[x] : x);
  }
  pull(x);
}
int fir(int x) {
  while (go[x][0]) push(x), x = go[x][0];
  return x;
}
void cyc(int c, int x) {
  int d = pd4(x);
  splay(c);
  push(x);
  int o = fir(go[c][1]);
  splay(o, c);
  go[c][1] = 0;
  go[c][d] = go[x][0];
  fa[go[c][d]] = c;
  go[x][0] = 0;
  rev(go[x][1]);
  go[o][1] = go[x][1];
  fa[go[o][1]] = o;
  go[x][1] = 0;
  go[o][0] = go[c][d ^ 1];
  fa[go[o][0]] = o;
  go[c][d ^ 1] = o;
  fa[o] = c;
  pull(o);
  pull(c);
}
void access(int x) {
  int y = 0, z = x;
  for (; x; y = x, x = fa[x]) {
    splay(x);
    if (fa[x] && go[fa[x]][2]) cyc(fa[x], x);
    go[x][1] = y;
    pull(x);
  }
  splay(z);
}
void mkrt(int x) {
  access(x);
  rev(x);
}
void path(int x, int y) {
  mkrt(x);
  access(y);
  splay(x, y);
}
int get(int x, int y) {
  if (x == y) return 0;
  path(x, y);
  if (fa[x] != y) return -1;
  return sum[y];
}
bool link(int x, int y, int w) {
  if (x == y) return false;
  path(x, y);
  if (fa[x] != y) {
    int e = make(w);
    fa[x] = e;
    fa[e] = y;
    return true;
  }
  if (hsc[y]) return false;
  int e = make(w), c = make(0);
  go[c][2] = go[x][1];
  fa[go[c][2]] = c;
  go[c][3] = e;
  fa[e] = c;
  go[x][1] = c;
  fa[c] = x;
  pull(c);
  pull(x);
  pull(y);
  return true;
}
bool cut(int x, int y, int w) {
  if (x == y) return false;
  path(x, y);
  if (fa[x] != y) return false;
  if (!isl(go[x][1])) return false;
  int c = go[x][1];
  push(c);
  if (go[c][2]) {
    int d = 0;
    if (isl(go[c][2]) && val[go[c][2]] == w) d = 2;
    if (isl(go[c][3]) && val[go[c][3]] == w) d = 3;
    if (!d) return false;
    go[x][1] = go[c][d ^ 1];
    fa[go[x][1]] = x;
    pull(x);
    pull(y);
    return true;
  } else {
    if (val[c] != w) return false;
    go[x][1] = 0;
    go[y][0] = 0;
    fa[x] = 0;
    pull(x);
    pull(y);
    return true;
  }
}
int n, m;
int U[500005], V[500005];
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  tsz = n;
  for (int i = 1; i <= m; i++) scanf("%d%d", &U[i], &V[i]);
  for (int i = 1, j = 0; i <= m; i++) {
    if (j < i) {
      j = i;
      link(U[j], V[j], 0);
    }
    while (j < m && link(U[j + 1], V[j + 1], 0)) j++;
    cut(U[i], V[i], 0);
    ans += j - i + 1;
  }
  printf("%lld\n", ans);
  return 0;
}

```