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 set1[100010 * 3], ql, qr;
long long sum[100010 * 3], v, ans[100010 * 3];
void insert(int L, int R, int o) {
  if (L == R) {
    set1[o] = v;
    sum[o] = 0;
    ans[o] = 0;
    return;
  }
  int mid = (L + R) >> 1;
  if (ql <= mid)
    insert(L, mid, o << 1);
  else
    insert(mid + 1, R, o << 1 | 1);
  ans[o] = sum[o] = 0;
  set1[o] = 0;
}
void pushdown(int L, int R, int o) {
  if (set1[o]) {
    int mid = (L + R) >> 1;
    ans[o << 1] += (mid - L + 1LL) * sum[o];
    ans[o << 1 | 1] += (R - mid) * sum[o];
    sum[o << 1] += sum[o];
    sum[o << 1 | 1] += sum[o];
    set1[o << 1] = set1[o << 1 | 1] = set1[o];
    set1[o] = sum[o] = 0;
  }
}
void pushup(int L, int R, int o) {
  if (set1[o << 1] == set1[o << 1 | 1])
    set1[o] = set1[o << 1];
  else
    set1[o] = 0;
  ans[o] = ans[o << 1] + ans[o << 1 | 1];
}
void update(int L, int R, int o) {
  if (ql <= L && qr >= R && set1[o]) {
    sum[o] += abs(set1[o] - v);
    ans[o] += (R - L + 1LL) * abs(set1[o] - v);
    set1[o] = v;
    return;
  }
  pushdown(L, R, o);
  int mid = (L + R) >> 1;
  if (ql <= mid) update(L, mid, o << 1);
  if (qr > mid) update(mid + 1, R, o << 1 | 1);
  pushup(L, R, o);
}
void find(int L, int R, int o) {
  if (ql <= L && R <= qr) {
    v += ans[o];
    return;
  }
  pushdown(L, R, o);
  int mid = (L + R) >> 1;
  if (ql <= mid) find(L, mid, o << 1);
  if (qr > mid) find(mid + 1, R, o << 1 | 1);
  pushup(L, R, o);
}
int main() {
  int n, m, i, j, k;
  while (scanf("%d%d", &n, &m) != EOF) {
    for (i = 1; i <= n; i++) {
      ql = i;
      v = i;
      insert(1, n, 1);
    }
    while (m--) {
      scanf("%d", &k);
      if (k == 1) {
        scanf("%d%d%I64d", &ql, &qr, &v);
        update(1, n, 1);
      } else {
        scanf("%d%d", &ql, &qr);
        v = 0;
        find(1, n, 1);
        printf("%I64d\n", v);
      }
    }
  }
  return 0;
}
