#include <bits/stdc++.h>
using namespace std;
struct SplayTree {
  struct Node {
    int ch[2] = {0, 0}, p = 0;
    bool flip = 0;
    int lazy = -1;
    int myvalue = -2;
    int pathvalue = -2;
    bool isedge = false;
  };
  vector<Node> T;
  SplayTree(int n) : T(n + 1) {}
  void push(int x) {
    if (!x) return;
    int l = T[x].ch[0], r = T[x].ch[1];
    if (T[x].flip) {
      T[l].flip ^= 1, T[r].flip ^= 1;
      swap(T[x].ch[0], T[x].ch[1]);
      T[x].flip = 0;
    }
    if (T[x].lazy != -1) {
      int val = T[x].lazy;
      if (T[x].isedge) {
        T[x].myvalue = val;
      }
      if (T[x].isedge || l || r) {
        T[x].pathvalue = val;
      }
      T[l].lazy = val;
      T[r].lazy = val;
      T[x].lazy = -1;
    }
  }
  void pull(int x) {
    int l = T[x].ch[0], r = T[x].ch[1];
    push(l);
    push(r);
    T[x].pathvalue = max({T[x].myvalue, T[l].pathvalue, T[r].pathvalue});
  }
  void set(int x, int d, int y) {
    T[x].ch[d] = y;
    T[y].p = x;
    pull(x);
  }
  void splay(int x) {
    auto dir = [&](int x) {
      int p = T[x].p;
      if (!p) return -1;
      return T[p].ch[0] == x ? 0 : T[p].ch[1] == x ? 1 : -1;
    };
    auto rotate = [&](int x) {
      int y = T[x].p, z = T[y].p, dx = dir(x), dy = dir(y);
      set(y, dx, T[x].ch[!dx]);
      set(x, !dx, y);
      if (~dy) set(z, dy, x);
      T[x].p = z;
    };
    for (push(x); ~dir(x);) {
      int y = T[x].p, z = T[y].p;
      push(z);
      push(y);
      push(x);
      int dx = dir(x), dy = dir(y);
      if (~dy) rotate(dx != dy ? x : y);
      rotate(x);
    }
  }
};
struct LinkCut : SplayTree {
  LinkCut(int n) : SplayTree(n) {}
  int access(int x) {
    int u = x, v = 0;
    for (; u; v = u, u = T[u].p) {
      splay(u);
      int& ov = T[u].ch[1];
      ov = v;
      pull(u);
    }
    return splay(x), v;
  }
  void reroot(int x) {
    access(x);
    T[x].flip ^= 1;
    push(x);
  }
  void link(int u, int v, int id) {
    assert(LCA(u, id) == 0);
    assert(LCA(v, id) == 0);
    Link(u, id);
    Link(v, id);
  }
  void Link(int u, int v) {
    u++;
    v++;
    reroot(u);
    access(v);
    T[u].p = v;
    pull(v);
  }
  void Cut(int u, int v) {
    u++;
    v++;
    reroot(u);
    access(v);
    T[v].ch[0] = T[u].p = 0;
    pull(v);
  }
  void cut(int u, int v, int id) {
    Cut(u, id);
    Cut(v, id);
  }
  int LCA(int u, int v) {
    u++;
    v++;
    if (u == v) return u;
    access(u);
    int ret = access(v);
    return T[u].p ? ret : 0;
  }
  int path(int u, int v) {
    u++;
    v++;
    reroot(u);
    access(v);
    return T[v].pathvalue;
  }
  int getValue(int i) { return path(i, i); }
  void setpath(int u, int v, int i) {
    u++;
    v++;
    reroot(u);
    access(v);
    T[v].lazy = i;
    push(v);
  }
  void uncycle(int u, int v, int i, int ei, int nu, int nv) {
    setpath(nu, nv, -2);
    cut(u, v, i);
    link(nu, nv, ei);
  }
};
int N, M;
pair<int, int> edges[500005];
int main() {
  cin >> N >> M;
  for (int i = 0; i < M; ++i) {
    cin >> edges[i].first >> edges[i].second;
    edges[i].first--;
    edges[i].second--;
  }
  LinkCut tree(N + M);
  for (int i = 0; i < M; ++i) {
    tree.T[N + i + 1].isedge = true;
  }
  auto insert = [&](int i) {
    auto [u, v] = edges[i];
    if (tree.LCA(u, v) == 0) {
      tree.link(u, v, N + i);
      return true;
    }
    int s = tree.path(u, v);
    if (s >= 0) return false;
    tree.setpath(u, v, i);
    return true;
  };
  auto erase = [&](int i) {
    auto [u, v] = edges[i];
    int s = tree.getValue(N + i);
    if (s < 0) {
      tree.cut(u, v, N + i);
      return;
    }
    auto [uu, vv] = edges[s];
    tree.uncycle(u, v, N + i, N + s, uu, vv);
  };
  long long ans = 0;
  for (int i = 0, j = 0; i < M; ++i) {
    while (j < M && insert(j)) j++;
    ans += j - i;
    erase(i);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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;
struct SplayTree {
  struct Node {
    int ch[2] = {0, 0}, p = 0;
    bool flip = 0;
    int lazy = -1;
    int myvalue = -2;
    int pathvalue = -2;
    bool isedge = false;
  };
  vector<Node> T;
  SplayTree(int n) : T(n + 1) {}
  void push(int x) {
    if (!x) return;
    int l = T[x].ch[0], r = T[x].ch[1];
    if (T[x].flip) {
      T[l].flip ^= 1, T[r].flip ^= 1;
      swap(T[x].ch[0], T[x].ch[1]);
      T[x].flip = 0;
    }
    if (T[x].lazy != -1) {
      int val = T[x].lazy;
      if (T[x].isedge) {
        T[x].myvalue = val;
      }
      if (T[x].isedge || l || r) {
        T[x].pathvalue = val;
      }
      T[l].lazy = val;
      T[r].lazy = val;
      T[x].lazy = -1;
    }
  }
  void pull(int x) {
    int l = T[x].ch[0], r = T[x].ch[1];
    push(l);
    push(r);
    T[x].pathvalue = max({T[x].myvalue, T[l].pathvalue, T[r].pathvalue});
  }
  void set(int x, int d, int y) {
    T[x].ch[d] = y;
    T[y].p = x;
    pull(x);
  }
  void splay(int x) {
    auto dir = [&](int x) {
      int p = T[x].p;
      if (!p) return -1;
      return T[p].ch[0] == x ? 0 : T[p].ch[1] == x ? 1 : -1;
    };
    auto rotate = [&](int x) {
      int y = T[x].p, z = T[y].p, dx = dir(x), dy = dir(y);
      set(y, dx, T[x].ch[!dx]);
      set(x, !dx, y);
      if (~dy) set(z, dy, x);
      T[x].p = z;
    };
    for (push(x); ~dir(x);) {
      int y = T[x].p, z = T[y].p;
      push(z);
      push(y);
      push(x);
      int dx = dir(x), dy = dir(y);
      if (~dy) rotate(dx != dy ? x : y);
      rotate(x);
    }
  }
};
struct LinkCut : SplayTree {
  LinkCut(int n) : SplayTree(n) {}
  int access(int x) {
    int u = x, v = 0;
    for (; u; v = u, u = T[u].p) {
      splay(u);
      int& ov = T[u].ch[1];
      ov = v;
      pull(u);
    }
    return splay(x), v;
  }
  void reroot(int x) {
    access(x);
    T[x].flip ^= 1;
    push(x);
  }
  void link(int u, int v, int id) {
    assert(LCA(u, id) == 0);
    assert(LCA(v, id) == 0);
    Link(u, id);
    Link(v, id);
  }
  void Link(int u, int v) {
    u++;
    v++;
    reroot(u);
    access(v);
    T[u].p = v;
    pull(v);
  }
  void Cut(int u, int v) {
    u++;
    v++;
    reroot(u);
    access(v);
    T[v].ch[0] = T[u].p = 0;
    pull(v);
  }
  void cut(int u, int v, int id) {
    Cut(u, id);
    Cut(v, id);
  }
  int LCA(int u, int v) {
    u++;
    v++;
    if (u == v) return u;
    access(u);
    int ret = access(v);
    return T[u].p ? ret : 0;
  }
  int path(int u, int v) {
    u++;
    v++;
    reroot(u);
    access(v);
    return T[v].pathvalue;
  }
  int getValue(int i) { return path(i, i); }
  void setpath(int u, int v, int i) {
    u++;
    v++;
    reroot(u);
    access(v);
    T[v].lazy = i;
    push(v);
  }
  void uncycle(int u, int v, int i, int ei, int nu, int nv) {
    setpath(nu, nv, -2);
    cut(u, v, i);
    link(nu, nv, ei);
  }
};
int N, M;
pair<int, int> edges[500005];
int main() {
  cin >> N >> M;
  for (int i = 0; i < M; ++i) {
    cin >> edges[i].first >> edges[i].second;
    edges[i].first--;
    edges[i].second--;
  }
  LinkCut tree(N + M);
  for (int i = 0; i < M; ++i) {
    tree.T[N + i + 1].isedge = true;
  }
  auto insert = [&](int i) {
    auto [u, v] = edges[i];
    if (tree.LCA(u, v) == 0) {
      tree.link(u, v, N + i);
      return true;
    }
    int s = tree.path(u, v);
    if (s >= 0) return false;
    tree.setpath(u, v, i);
    return true;
  };
  auto erase = [&](int i) {
    auto [u, v] = edges[i];
    int s = tree.getValue(N + i);
    if (s < 0) {
      tree.cut(u, v, N + i);
      return;
    }
    auto [uu, vv] = edges[s];
    tree.uncycle(u, v, N + i, N + s, uu, vv);
  };
  long long ans = 0;
  for (int i = 0, j = 0; i < M; ++i) {
    while (j < M && insert(j)) j++;
    ans += j - i;
    erase(i);
  }
  cout << ans << endl;
  return 0;
}

```