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;
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
const int N = 100002;
int n, col[4 * N];
long long tree[4 * N], lazy[4 * N];
void build(int node, int s, int e) {
  if (s > e) return;
  if (s == e) {
    col[node] = s;
    return;
  }
  int mid = (s + e) / 2;
  build((2 * (node)), s, mid);
  build((2 * (node) + 1), mid + 1, e);
}
void relax(int node, int s, int e) {
  int mid = (s + e) / 2;
  if (lazy[node] > 0) {
    tree[(2 * (node))] += lazy[node] * (mid - s + 1);
    lazy[(2 * (node))] += lazy[node];
    tree[(2 * (node) + 1)] += lazy[node] * (e - mid);
    lazy[(2 * (node) + 1)] += lazy[node];
    lazy[node] = 0;
  }
  if (col[node] > 0) {
    col[(2 * (node))] = col[node];
    col[(2 * (node) + 1)] = col[node];
  }
}
void traverse(int node, int s, int e, int c) {
  if (s == e || col[node] > 0) {
    tree[node] += abs(col[node] - c) * 1LL * (e - s + 1);
    lazy[node] += abs(col[node] - c);
    col[node] = c;
    return;
  }
  relax(node, s, e);
  int mid = (s + e) / 2;
  traverse((2 * (node)), s, mid, c);
  traverse((2 * (node) + 1), mid + 1, e, c);
  tree[node] = tree[(2 * (node))] + tree[(2 * (node) + 1)];
  if (col[(2 * (node))] > 0 && col[(2 * (node) + 1)] == col[(2 * (node))])
    col[node] = col[(2 * (node))];
  else
    col[node] = 0;
}
void update(int node, int s, int e, int lo, int hi, int val) {
  if (s > e || s > hi || lo > e) return;
  if (s >= lo && e <= hi) {
    traverse(node, s, e, val);
    return;
  }
  relax(node, s, e);
  int mid = (s + e) / 2;
  update((2 * (node)), s, mid, lo, hi, val);
  update((2 * (node) + 1), mid + 1, e, lo, hi, val);
  tree[node] = tree[(2 * (node))] + tree[(2 * (node) + 1)];
  if (col[(2 * (node))] > 0 && col[(2 * (node) + 1)] == col[(2 * (node))])
    col[node] = col[(2 * (node))];
  else
    col[node] = 0;
}
long long query(int node, int s, int e, int lo, int hi) {
  if (s > e || lo > e || s > hi) return 0;
  if (s >= lo && e <= hi) return tree[node];
  relax(node, s, e);
  int mid = (s + e) / 2;
  long long a = query((2 * (node)), s, mid, lo, hi);
  long long b = query((2 * (node) + 1), mid + 1, e, lo, hi);
  return a + b;
}
int main() {
  int q;
  scanf("%d%d", &n, &q);
  build(1, 1, n);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l, r, val;
      scanf("%d%d%d", &l, &r, &val);
      update(1, 1, n, l, r, val);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      printf("%lld\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}
