Farmer John's $N$ $(1 \leq N \leq 2 \cdot 10^5)$ cows are lined up in a circle such that for each $i$ in $1,2,\dots,N-1$, the cow to the right of cow $i$ is cow $i+1$, and the cow to the right of cow $N$ is cow $1$. The $i$th cow has a bucket with integer capacity $a_i$ $(1 \leq a_i \leq 10^9)$ liters. All buckets are initially full.

Every minute, the cows exchange milk according to a string $s_1s_2\dots s_N$ consisting solely of the characters $\text{‘L’}$ and $\text{‘R’}$. if the $i$th cow has at least $1$ liter of milk, she will pass $1$ liter of milk to the cow to her left if $s_i=\text{‘L’}$, or to the right if $s_i=\text{‘R’}$. All exchanges happen simultaneously (i.e., if a cow has a full bucket but gives away a liter of milk but also receives a liter, her milk is preserved). If a cow's total milk ever ends up exceeding $a_i$, then the excess milk will be lost.

FJ wants to know: after $M$ minutes $(1 \leq M \leq 10^9$), what is the total amount of milk left among all cows?

The first line contains $N$ and $M$.

The second line contains a string $s_1s_2\dots s_N$ consisting solely of the characters $\text{‘L’}$ or $\text{‘R’}$, denoting the direction each cow will pass their milk in.

The third line contains integers $a_1, a_2, \dots, a_N$, the capacities of each bucket.

Output an integer, the sum of milk among all cows after $M$ minutes.

Note that the large size of integers involved in this problem may require the use of 64-bit integer data types (e.g., a "long long" in C/C++).

• Inputs 4-8: $N,M \le 1000$
• Inputs 9-16: No additional constraints.

2

Cows $2$ and $3$ pass each other one liter of milk, so their milk is preserved. When cow $1$ passes their milk to cow $2$, cow $2$'s bucket overflows, and one liter of milk is lost after one minute.

14

Each cow is passing a liter of milk to the cow on the left and gaining a liter of milk from the cow on the right, so all of the milk is preserved regardless of how much time passes.

38

Initially, there are a total of 51 liters of milk. After 5 minutes, cows $3$, $6$, and $7$ will lose 5, 3, and 5 liters of milk respectively. Therefore, a total of 38 liters of milk remain.