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 double pi = acos(-1.0);
const double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const int maxn = 100000 + 10;
long long sum[maxn << 2], col[maxn << 2], add[maxn << 2];
int n, m;
void build(int l, int r, int rt) {
  sum[rt] = add[rt] = 0;
  if (l == r) {
    col[rt] = l;
    return;
  }
  col[rt] = 0;
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
}
void pushUp(int rt) {
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
  if (col[rt << 1] == col[rt << 1 | 1])
    col[rt] = col[rt << 1];
  else
    col[rt] = 0;
}
void Change(int l, int r, int rt, int x) {
  if (col[rt]) {
    add[rt] += abs(col[rt] - x);
    sum[rt] += 1LL * abs(col[rt] - x) * (r - l + 1);
    col[rt] = x;
    return;
  }
  int m = (l + r) >> 1;
  Change(l, m, rt << 1, x);
  Change(m + 1, r, rt << 1 | 1, x);
  pushUp(rt);
}
void pushDown(int rt, int m) {
  if (add[rt]) {
    add[rt << 1] += add[rt];
    add[rt << 1 | 1] += add[rt];
    sum[rt << 1] += (m - (m >> 1)) * add[rt];
    sum[rt << 1 | 1] += (m >> 1) * add[rt];
    col[rt << 1] = col[rt << 1 | 1] = col[rt];
    add[rt] = 0;
  }
}
void Update(int L, int R, int x, int l, int r, int rt) {
  if (L <= l && R >= r) {
    Change(l, r, rt, x);
    return;
  }
  pushDown(rt, r - l + 1);
  int m = (l + r) >> 1;
  if (L <= m) Update(L, R, x, l, m, rt << 1);
  if (R > m) Update(L, R, x, m + 1, r, rt << 1 | 1);
  pushUp(rt);
}
long long Query(int L, int R, int l, int r, int rt) {
  if (L <= l && R >= r) {
    return sum[rt];
  }
  pushDown(rt, r - l + 1);
  int m = (l + r) >> 1;
  long long ret = 0;
  if (L <= m) ret += Query(L, R, l, m, rt << 1);
  if (R > m) ret += Query(L, R, m + 1, r, rt << 1 | 1);
  return ret;
}
int main() {
  while (scanf("%d%d", &n, &m) == 2) {
    build(1, n, 1);
    while (m--) {
      int op, l, r, x;
      scanf("%d", &op);
      if (op == 1) {
        scanf("%d%d%d", &l, &r, &x);
        Update(l, r, x, 1, n, 1);
      } else if (op == 2) {
        scanf("%d%d", &l, &r);
        printf("%I64d\n", Query(l, r, 1, n, 1));
      }
    }
  }
  return 0;
}
