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;
int N, M;
struct Bit {
  vector<long long int> tree, lazy;
  vector<int> same;
  void init(int l, int r, int n) {
    if (l == r) {
      tree[n] = 0;
      lazy[n] = 0;
      same[n] = l;
      return;
    }
    int m = (l + r) >> 1;
    init(l, m, 2 * n);
    init(m + 1, r, 2 * n + 1);
    tree[n] = tree[2 * n] + tree[2 * n + 1];
    same[n] = same[2 * n] == same[2 * n + 1] ? same[2 * n] : -1;
  }
  void prop(int l, int r, int n) {
    if (l != r) {
      int m = (l + r) >> 1;
      tree[2 * n] += (long long int)(m - l + 1) * lazy[n];
      lazy[2 * n] += lazy[n];
      if (same[n] != -1) same[2 * n] = same[n];
      tree[2 * n + 1] += (long long int)(r - m) * lazy[n];
      lazy[2 * n + 1] += lazy[n];
      if (same[n] != -1) same[2 * n + 1] = same[n];
    }
    lazy[n] = 0;
  }
  void upd(int a, int b, int c, int l, int r, int n) {
    if (b < l || r < a) return;
    if (a <= l && r <= b && same[n] != -1) {
      tree[n] += (long long int)(r - l + 1) * abs(same[n] - c);
      lazy[n] = abs(same[n] - c);
      same[n] = c;
      return;
    }
    int m = (l + r) >> 1;
    prop(l, m, 2 * n);
    prop(m + 1, r, 2 * n + 1);
    prop(l, r, n);
    upd(a, b, c, l, m, 2 * n);
    upd(a, b, c, m + 1, r, 2 * n + 1);
    tree[n] = tree[2 * n] + tree[2 * n + 1];
    same[n] = same[2 * n] == same[2 * n + 1] ? same[2 * n] : -1;
  }
  long long int qur(int a, int b, int l, int r, int n) {
    if (b < l || r < a) return 0;
    if (a <= l && r <= b) return tree[n];
    prop(l, r, n);
    int m = (l + r) >> 1;
    long long int L = qur(a, b, l, m, 2 * n);
    long long int R = qur(a, b, m + 1, r, 2 * n + 1);
    return L + R;
  }
} bit;
int main() {
  scanf("%d %d", &N, &M);
  bit.tree.resize(4 * N);
  bit.lazy.resize(4 * N);
  bit.same.resize(4 * N);
  bit.init(0, N - 1, 1);
  for (int i = 0; i < M; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int l, r, x;
      scanf("%d %d %d", &l, &r, &x);
      l--;
      r--;
      x--;
      bit.upd(l, r, x, 0, N - 1, 1);
    }
    if (t == 2) {
      int l, r;
      scanf("%d %d", &l, &r);
      l--;
      r--;
      printf("%I64d\n", bit.qur(l, r, 0, N - 1, 1));
    }
  }
}
