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;
const int N = 100005;
int n, q;
long long sum[N << 2], delta[N << 2], mark[N << 2];
void init() {
  memset(sum, 0, sizeof(sum));
  memset(delta, 0, sizeof(delta));
  memset(mark, 0, sizeof(mark));
}
void build(int l, int r, int rt) {
  if (l == r) {
    mark[rt] = l;
  } else {
    int m = (l + r) >> 1;
    build(l, m, rt << 1);
    build(m + 1, r, rt << 1 | 1);
    mark[rt] = 0;
  }
}
void pushdown(long long x, int l, int r, int rt) {
  if (mark[rt] > 0) {
    delta[rt] += (long long)abs(mark[rt] - x);
    sum[rt] += (long long)abs(mark[rt] - x) * (long long)(r - l + 1);
    mark[rt] = -1;
  } else {
    int m = (l + r) >> 1;
    pushdown(x, l, m, rt << 1);
    pushdown(x, m + 1, r, rt << 1 | 1);
    sum[rt] =
        sum[rt << 1] + sum[rt << 1 | 1] + delta[rt] * (long long)(r - l + 1);
  }
}
void update(int L, int R, long long x, int l, int r, int rt) {
  if (L <= l && r <= R) {
    pushdown(x, l, r, rt);
    mark[rt] = x;
  } else {
    if (mark[rt] > 0)
      mark[rt << 1] = mark[rt << 1 | 1] = mark[rt], mark[rt] = 0;
    int m = (l + r) >> 1;
    if (m >= L) update(L, R, x, l, m, rt << 1);
    if (m < R) update(L, R, x, m + 1, r, rt << 1 | 1);
    sum[rt] =
        sum[rt << 1] + sum[rt << 1 | 1] + delta[rt] * (long long)(r - l + 1);
  }
}
long long qurry(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) return sum[rt];
  long long ret = 0;
  int m = (l + r) >> 1;
  if (m >= L) ret += qurry(L, R, l, m, rt << 1);
  if (m < R) ret += qurry(L, R, m + 1, r, rt << 1 | 1);
  return ret + delta[rt] * (long long)(min(r, R) - max(L, l) + 1);
}
int main() {
  ios::sync_with_stdio(false);
  while (scanf("%d %d", &n, &q) != EOF) {
    init();
    build(1, n, 1);
    int op, a, b, c;
    while (q--) {
      scanf("%d %d %d", &op, &a, &b);
      if (op == 1) {
        scanf("%d", &c);
        update(a, b, c, 1, n, 1);
      } else {
        printf("%I64d\n", qurry(a, b, 1, n, 1));
      }
    }
  }
  return 0;
}
