DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

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


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
struct ST {
  struct node {
    T sum = 0, lzy = 0;
  };
  vector<node> t;
  ST(int n) : t(vector<node>(n << 2)) {}
  void push(int v, int l, int r) {
    t[v].sum += t[v].lzy * (r - l + 1);
    if (l ^ r) {
      t[v * 2].lzy += t[v].lzy;
      t[v * 2 + 1].lzy += t[v].lzy;
    }
    t[v].lzy = 0;
  }
  void modify(int l, int r, int L, int R, int x, int v = 1) {
    push(v, l, r);
    if (l > R || r < L) return;
    if (l >= L && r <= R) {
      t[v].lzy += x;
      push(v, l, r);
      return;
    }
    modify(l, (l + r) / 2, L, R, x, v * 2);
    modify((l + r) / 2 + 1, r, L, R, x, v * 2 + 1);
    t[v].sum = t[v * 2].sum + t[v * 2 + 1].sum;
  }
  T query(int l, int r, int L, int R, int v = 1) {
    push(v, l, r);
    if (l > R || r < L) return 0;
    if (l >= L && r <= R) {
      return t[v].sum;
    }
    return query(l, (l + r) / 2, L, R, v * 2) +
           query((l + r) / 2 + 1, r, L, R, v * 2 + 1);
  }
};
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  set<tuple<int, int, int> > st;
  for (int i = 1; i <= n; i++) {
    st.insert(make_tuple(i, i, i));
  }
  ST<long long> tr(n + 1);
  for (int i = 1; i <= m; i++) {
    int typ;
    cin >> typ;
    if (typ == 1) {
      int l, r, x;
      cin >> l >> r >> x;
      vector<tuple<int, int, int> > v;
      for (auto it = st.lower_bound(make_tuple(l, -1, -1)); it != st.end();
           ++it) {
        v.push_back(*it);
        auto [R, L, X] = *it;
        if (R > r) break;
      }
      for (auto y : v) {
        st.erase(y);
        auto [R, L, X] = y;
        if (L < l) {
          st.insert(make_tuple(l - 1, L, X));
        }
        if (R > r) {
          st.insert(make_tuple(R, r + 1, X));
        }
        R = min(r, R);
        L = max(L, l);
        tr.modify(1, n, L, R, abs(x - X));
      }
      st.insert(make_tuple(r, l, x));
    } else {
      int l, r;
      cin >> l >> r;
      cout << tr.query(1, n, l, r) << "\n";
    }
  }
}
