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 maxn = 5 * 1e5;
long long s[maxn], del[maxn], mark[maxn];
void build(int u, int l, int r) {
  if (l < r) {
    int mid = (l + r) / 2;
    build(u * 2, l, mid), build(u * 2 + 1, mid + 1, r);
  } else
    mark[u] = l;
}
long long query(int u, int l, int r, int x, int y) {
  if (y < l || x > r) return 0;
  if (x <= l && r <= y) return s[u];
  int mid = (l + r) / 2;
  long long L = query(u * 2, l, mid, x, y);
  long long R = query(u * 2 + 1, mid + 1, r, x, y);
  return L + R + max(0, min(r, y) - max(l, x) + 1) * del[u];
}
void clear(int u, int l, int r, int z) {
  if (mark[u]) {
    del[u] += abs(mark[u] - z);
    s[u] += 1LL * (r - l + 1) * abs(mark[u] - z);
    mark[u] = 0;
  } else {
    if (l == r) return;
    int mid = (l + r) / 2, lc = u * 2, rc = u * 2 + 1;
    clear(lc, l, mid, z);
    clear(rc, mid + 1, r, z);
    s[u] = s[lc] + s[rc] + 1LL * (r - l + 1) * del[u];
  }
}
void modify(int u, int l, int r, int x, int y, int z) {
  if (y < l || x > r) return;
  if (x <= l && r <= y) {
    clear(u, l, r, z);
    mark[u] = z;
    return;
  }
  int mid = (l + r) / 2, lc = u * 2, rc = u * 2 + 1;
  if (mark[u]) mark[lc] = mark[rc] = mark[u], mark[u] = 0;
  modify(lc, l, mid, x, y, z);
  modify(rc, mid + 1, r, x, y, z);
  s[u] = s[lc] + s[rc] + 1LL * (r - l + 1) * del[u];
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  int type, x, y, z;
  ;
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &type, &x, &y);
    if (type == 1) {
      scanf("%d", &z);
      modify(1, 1, n, x, y, z);
    } else
      (cout << query(1, 1, n, x, y) << endl);
  }
  return 0;
}
